Yes, for accurate AC calculations, the power factor (typically 0.8) must be included.

Yes, multiply kW by 1000 to get watts, then apply the formula.

It provides highly accurate results, using standard electrical formulas.

Yes, it works for DC, single-phase, and three-phase AC systems.

Use 0.8 as a common default for general calculations.

For DC: 1000 ÷ 120 = 8.33 Amps.

Yes, higher voltage results in lower current for the same power.

Absolutely, it’s ideal for off-grid, solar, and battery applications.

Yes, it’s responsive and works on all smartphones and devices.

For DC: Amps = Watts ÷ Volts. For AC, you also consider power factor and phase.

Yes, for accurate AC calculations, the power factor (typically 0.8) must be included.

Yes, multiply kW by 1000 to get watts, then apply the formula.

It provides highly accurate results, using standard electrical formulas.

Yes, it works for DC, single-phase, and three-phase AC systems.

Use 0.8 as a common default for general calculations.

For DC: 1000 ÷ 120 = 8.33 Amps.

Yes, higher voltage results in lower current for the same power.

Absolutely, it’s ideal for off-grid, solar, and battery applications.

Yes, it’s responsive and works on all smartphones and devices.

Volts = Watts ÷ Amps (for DC)

For single-phase: Volts = Watts ÷ (Amps × Power Factor)

Use: Volts = Watts ÷ (√3 × Amps × Power Factor)

Most systems use 0.8 as a general assumption.

Yes, it works for DC and both AC systems.

Yes, it's ideal for panels, inverters, and load estimations.

The result is in volts (V).

Yes. Use: Watts = Volts × Amps (× Power Factor if AC).

Use 0.8 for general estimates or consult device specs.

Yes! It’s responsive and optimized for all devices.

Watt-hour is a unit of energy representing power consumption over time.

Multiply watts by the number of hours the device runs.

No, watts measure power; watt-hours measure energy used over time.

Yes, watt-hours apply to all electrical devices.

It helps estimate battery capacity and runtime.

Use Watt-Hours = Watts × Hours.

No, this is theoretical energy use without losses.

Yes, it helps estimate daily or monthly energy output.

Use decimal values, e.g., 0.5 hours for 30 minutes.

Yes, fully responsive on all devices.

A joule is a unit of energy; it equals one watt of power used for one second.

Multiply watts by time in seconds (Joules = Watts × Seconds).

No, watts measure power; joules measure energy.

Watts = energy per unit time; Joules = total energy.

Yes, it helps estimate energy output over time.

Absolutely. It's commonly used in mechanics and thermodynamics.

You’ll need to convert time into seconds first.

The joule (J) is the standard SI unit for energy.

Yes, joules apply to energy, regardless of current type.

Yes, the calculator and layout are fully responsive.

The formula is: kVA = Watts ÷ (1000 × Power Factor).

Power factor adjusts for the phase difference between voltage and current in AC systems.

A typical default value is 0.8 for industrial and commercial loads.

Yes, it works for both, just ensure the correct power factor is used.

kVA stands for kilovolt-amperes, the unit of apparent power.

Yes, if you know the power factor: kVA = kW ÷ Power Factor.

Yes, it’s fully responsive and mobile-friendly.

Yes, it ranges between 0 and 1 in real-world AC circuits.

To size generators or UPS systems, which are often rated in kVA.

Absolutely, it’s 100% free with no signup required.

A kilowatt-hour is a unit of energy equal to using 1000 watts for one hour.

Multiply watts by hours used and divide by 1000. Formula: kWh = (W × H) / 1000.

It helps track energy consumption and estimate electricity bills.

Yes, it's perfect for calculating energy used by TVs, ACs, ovens, etc.

Not exactly. 1 kWh means 1000 watts used continuously for 1 hour.

kW is a unit of power; kWh is a unit of energy (power × time).

Yes, if you enter correct wattage and usage time, it gives precise results.

Yes. Multiply daily kWh usage by the number of days in a month.

Yes, they use fewer watts and thus consume less energy over time.

Absolutely. By knowing usage, you can adjust habits to save energy.

VA stands for volt-ampere, a unit that measures apparent power in an electrical circuit. It’s the product of voltage (V) and current (A).

To convert amps to VA, multiply the current (in amps) by the voltage (in volts). The formula is: VA = Volts × Amps.

Not exactly. VA represents apparent power, while watts represent real power. In purely resistive loads, they can be equal (when power factor = 1).

VA accounts for both real and reactive power in AC systems. It's important for sizing transformers, UPS systems, and power supplies.

Apparent power is the combination of real power (watts) and reactive power (VAR). It’s measured in volt-amperes (VA).

No. Power factor is needed when converting VA to watts. But to get VA from amps, you only need current and voltage.

This calculator is mainly for AC circuits where apparent power is relevant. In DC circuits, watts = volts × amps, with no reactive power.

This calculator is designed for single-phase systems. For 3-phase systems, use the formula: VA = √3 × Volts × Amps.

Enter the voltage used by your equipment or supply source. Common values include 120V, 230V, or 240V depending on your region.

UPS systems are rated in VA. Knowing your total VA requirement helps ensure UPS can handle the full electrical load safely.

Volts = Watts ÷ Amps. For 3-phase, divide by √3 × Amps.

Yes, power in watts is necessary to calculate voltage from current.

AC voltage alternates direction; DC voltage flows in one direction.

Voltage between any two of the three phase conductors.

Yes, the formula Volts = Watts ÷ Amps works for DC as well.

Yes, 3-phase uses √3 in the denominator for line-to-line voltage.

√3 (approx. 1.732) accounts for phase angle differences in 3-phase systems.

It provides accurate results if input values are correct.

Yes, it's lower by a factor of √3 in a balanced 3-phase system.

Yes, you can generate a widget using our embeddable tool.

Watts = Amps × Volts for DC; AC may include a power factor.

Yes, in AC circuits, power factor is crucial to get accurate watt values.

Yes. Three-phase uses √3 × Amps × Volts × PF.

Yes, it supports DC, single-phase, and three-phase systems.

Amps (A) for current and Volts (V) for voltage.

It's the efficiency ratio between real power and apparent power in AC circuits.

To avoid errors and save time on repetitive calculations.

Most household devices use single-phase electricity.

You need to know voltage too; then multiply it by amps.

Yes, it uses standard formulas used in electrical engineering.

kVA stands for kilovolt-ampere, a unit of apparent power.

Multiply amps × voltage × √3 (for 3-phase) or × voltage (for single-phase), then divide by 1000.

kVA = (Amps × Volts) / 1000.

kVA = (√3 × Amps × Volts) / 1000.

√3 accounts for the phase difference between voltages in a 3-phase system.

Yes, it supports both single-phase and 3-phase conversions.

You need to input current (Amps) and voltage (Volts).

Yes, it uses industry-standard electrical formulas.

Apparent power is the product of current and voltage without considering power factor.

No, power factor is not used when calculating apparent power in kVA.

kW = (V × I × PF) / 1000 for single phase.

Power factor (PF) is the ratio of real power to apparent power, typically between 0.8 and 1.

No, power factor is only relevant for AC circuits.

Use kW = (√3 × V × I × PF) / 1000.

Yes, it's based on standard electrical formulas.

Yes, as long as voltage and current are known.

Amps, Volts, Power Factor, and kW.

Yes, unless the load is purely resistive (PF=1).

Yes, by rearranging the formula: Amps = (kW × 1000) / (V × PF).

No, frequency is not considered in basic kW conversion.

VA stands for volt-ampere, a unit of apparent power in an electrical circuit.

Watts = VA × Power Factor.

Power factor is the ratio of real power to apparent power in a circuit, ranging from 0 to 1.

Multiply the VA value by the power factor: Watts = VA × PF.

Only if the power factor is 1, which occurs in purely resistive loads.

VA is apparent power; watts is real power actually used by a device.

Yes, if you account for line voltage and the correct power factor.

Use an estimated PF (e.g., 0.8 for inductive loads) or consult equipment specifications.

It determines how much of the apparent power is converted into usable real power.

Electricity bills are usually based on watts, not VA, unless you're billed for apparent power too.

It is the process of converting volt-amperes (VA) to kilovolt-amperes (kVA) by dividing VA by 1,000.

The formula is: kVA = VA ÷ 1000.

Converting to kVA is useful when dealing with large power ratings in electrical systems.

No, VA is volt-amperes and kVA is kilovolt-amperes. 1 kVA = 1,000 VA.

Yes, the calculator works for both single-phase and three-phase systems.

kW measures real power while kVA measures apparent power.

No, for VA to kVA conversion, power factor is not needed.

2000 VA is equal to 2 kVA.

Yes, generator sizes are often rated in kVA.

Yes, multiply the kVA value by 1,000 to get VA.

Divide the VA value by voltage: Amps = VA ÷ V.

Yes, voltage is required to calculate current.

VA stands for volt-amperes, representing apparent power.

Watts represent real power; VA is apparent power.

No, VA to amps doesn't consider power factor.

Use I = VA ÷ (√3 × V) for line-to-line voltage.

Yes, simply use I = VA ÷ V for DC too.

Current increases, which can overheat wires.

Yes, it's based on standard electrical formulas.

Yes, for sizing generators based on VA rating.

Use the formula: kW = (VA × Power Factor) ÷ 1000.

Power factor is the ratio of real power to apparent power, usually between 0.7 and 1.

Yes, it directly affects the kW result from VA.

VA (volt-ampere) is the unit of apparent power in AC systems.

No, power factor is necessary for accurate conversion.

Use the formula: kW = (VA × PF × √3) ÷ 1000.

kW stands for kilowatt, a unit of real power.

For DC, VA = W, so no PF is needed; divide by 1000 to get kW.

Because VA includes both real and reactive power, not just real power.

Yes, it's accurate and supports both single and three-phase loads.

Multiply volts by amps. For AC, also multiply by the power factor and √3 for 3-phase.

Watts = Volts × Amps.

No, power factor is only used in AC circuits.

It’s a number (0–1) that shows how efficiently electrical power is used in AC circuits.

Use Watts = Volts × Amps × Power Factor.

Use Watts = √3 × Volts × Amps × PF.

Yes, it supports DC, single-phase, and three-phase systems.

√3 ≈ 1.732, used in 3-phase AC power calculations.

Yes, especially for calculating power usage based on voltage and current.

Yes, it's helpful when calculating wattage output based on system voltage and current.

Amps = Watts ÷ Volts.

Yes, you need both watts and volts to calculate amps.

Yes, it's suitable for both DC and single-phase AC circuits.

Yes, for 3-phase: Amps = Watts ÷ (√3 × Volts × PF).

Power factor (PF) is the ratio of real power to apparent power in AC circuits.

No, watts are required for an accurate conversion.

Yes, it helps electricians estimate current values quickly.

It is accurate if you input correct wattage and voltage values.

Yes, you can use it to convert DC voltage and power to current.

Yes, this volts to amps calculator is 100% free.

J = 0.5 × C × V², where C is capacitance in farads and V is voltage in volts.

No, capacitance is required for accurate joules calculation using voltage.

A joule is a unit of energy equal to the work done by one ampere passing through one ohm for one second.

Voltage in volts and capacitance in farads.

Yes, it's commonly used to calculate stored energy in capacitors.

Yes, just convert them to farads: 1μF = 1e-6 F.

It is generally used for DC circuits or peak AC values.

J = 0.5 × 0.01 × 10² = 0.5 Joules.

Yes, it's commonly used for electrical physics problems.

Joules will increase by a factor of four, since energy is proportional to voltage squared.

An electron volt is the energy gained by one electron moving through a potential difference of one volt.

Multiply volts by the elementary charge (1.602×10⁻¹⁹ C).

Yes, it's a linear conversion using the charge of an electron.

Used in physics, electronics, and quantum mechanics.

The charge of one electron: 1.602 × 10⁻¹⁹ coulombs.

No, eV is a unit of energy, not voltage.

Yes, it’s suitable for academic and lab work.

It’s highly accurate using a standard constant value.

Yes, it’s 100% free and easy to use.

Not in basic conversions, but it may in physical experiments.

Use the formula: kW = (V × I × PF) ÷ 1000 for single-phase or include √3 for three-phase.

kW = (√3 × V × I × PF) ÷ 1000.

PF stands for power factor, which represents efficiency.

No, you need both volts and amps to calculate power.

Yes, it supports 3-phase systems with √3 included.

Most systems have a PF between 0.8 and 1.

kW is real power; kVA is apparent power.

Yes, it's accurate for industrial and commercial use.

Yes, specify whether it’s single or three-phase.

Yes, it's completely free and easy to use.

Use the formula: Watts = Watt-hours ÷ Hours.

A watt-hour (Wh) is a unit of energy representing one watt for one hour.

Yes, you need the time (in hours) to convert Wh to W.

Yes, it's commonly used to calculate battery discharge power.

Yes, it helps estimate power from stored solar energy.

Wh is energy, W is power (rate of energy use).

Yes, the basic formula applies universally.

You can measure or estimate how long the device runs.

Yes, it's 100% free and available online.

No, but you can use the inverse formula: Wh = W × h.

Use the formula: mAh = (Wh × 1000) ÷ Voltage (V).

Milliamp-hours (mAh) measure electric charge or battery capacity.

You need watt-hours (Wh) and voltage (V).

Yes, it's commonly used for lithium-ion cells.

Yes, use Wh = (mAh × V) ÷ 1000.

To convert amps to milliamps since 1 A = 1000 mA.

Use 5V unless your device specifies otherwise.

Yes, it's free and accessible online.

Yes, it helps determine battery capacity needed.

The result is in milliamp-hours (mAh).

Use kW = Wh ÷ 1000 ÷ time in hours.

A kilowatt is 1000 watts, a unit of power.

Watt-hours measure the total energy consumed.

Yes, because power (kW) = energy ÷ time.

No, kW is a unit of power, not energy.

Wh is energy, kW is power.

Only if time = 1 hour, otherwise time is required.

Yes, it's completely free and online.

To convert watts to kilowatts.

Used in energy audits, electrical loads, and appliance ratings.

Use the formula V = J ÷ C, where J is energy in joules and C is charge in coulombs.

V = J / C

Coulombs (C)

Yes, it's useful for energy stored in capacitors.

Volt is the unit of electric potential or voltage.

A joule is a unit of energy.

No, charge (C) is required to compute voltage from energy.

Yes, it's a free online tool.

Common in physics, electronics, and engineering.

Joules measure energy; volts measure electric potential.

Use the formula kW = (J ÷ t) ÷ 1000, where J is energy in joules and t is time in seconds.

kW = (J / seconds) / 1000.

Time must be in seconds.

Yes, using time, energy can be converted to power (kW).

Yes, 1 kW = 1000 watts.

Yes, especially in energy and power conversion.

Convert minutes to seconds first (1 minute = 60 seconds).

Kilowatt is a unit of power equal to 1000 watts.

Yes, it’s 100% free.

Used in science, engineering, and power systems.

Use the formula Wh = (mAh × V) ÷ 1000.

Wh = (mAh × Voltage) ÷ 1000.

It stands for milliamp-hours, indicating battery capacity.

Use the battery's rated voltage, usually printed on the label.

Yes, (5000 × 5) ÷ 1000 = 25 Wh.

Yes, completely free to use online.

Yes, just enter the typical 3.6V or 3.7V.

To convert milliamp-hours to amp-hours for the watt-hour equation.

Yes, it helps compare real energy capacity.

In electronics, power banks, and battery sizing.

Multiply kVA by the power factor: kW = kVA × PF.

Power factor is the ratio of real power to apparent power, ranging from 0 to 1.

kW = kVA × Power Factor.

kVA stands for kilovolt-amps and represents apparent power.

Yes, if the load is balanced and PF is known.

It’s usually provided by equipment specs or measured using meters.

Because kW is real power, affected by the power factor.

Yes, it's a 100% free calculator.

Power factor ranges from 0 (worst) to 1 (ideal).

Yes, divide kW by PF: kVA = kW ÷ PF.

Multiply kVA by 1000. VA = kVA × 1000.

VA = kVA × 1000.

VA (Volt-Amps) represents apparent power.

kVA is kilovolt-amps; VA is volt-amps. 1 kVA = 1000 VA.

kVA is used for larger power ratings.

Yes, the formula works for all AC systems.

Yes, by definition.

Yes, divide VA by 1000.

No. VA is apparent power, watts is real power.

Yes, this calculator is 100% free.

Multiply kVA by power factor and 1000: Watts = kVA × PF × 1000.

Ratio of real power to apparent power, between 0 and 1.

Watts = kVA × Power Factor × 1000.

Kilovolt-amps, a measure of apparent power.

Yes, watts is real power, usually less due to PF.

Yes, if PF and kVA are known.

To convert kVA (thousands of VA) to VA units.

Yes, free and easy to use.

Yes, kVA = watts ÷ (PF × 1000).

In electrical load and power system design.

Use the formula: Amps = (kW × 1000) ÷ (V × PF) for single-phase or divide by (√3 × V × PF) for three-phase.

Amps = (kW × 1000) ÷ (Voltage × Power Factor).

Amps = (kW × 1000) ÷ (√3 × Voltage × Power Factor).

It's the ratio of real power to apparent power, usually between 0.8 and 1.0.

Voltage is needed because current depends on both voltage and power.

It is for AC systems — single-phase and three-phase.

Yes, especially when PF is known or assumed.

Very accurate if PF and voltage are correct.

Power in kilowatts (kW), voltage in volts (V), current in amps (A).

Yes, this calculator is 100% free to use online.

A joule is the SI unit of energy, equal to one watt-second.

Multiply kilowatts by 1000 and then by time in seconds: J = kW × 1000 × s.

Energy (J) = Power (kW) × 1000 × Time (s).

To quickly find how much energy is used or produced over time.

Yes, it uses the standard physical formula for energy conversion.

Convert minutes or hours to seconds before using the calculator.

Engineering, physics, energy audits, and electrical design rely on it.

Yes, the calculator is fully responsive and mobile-friendly.

No installation is needed—just use it online for free.

The result is in joules; divide by 1000 for kilojoules.

kVA = kW / Power Factor (PF).

Power Factor is the ratio of real power (kW) to apparent power (kVA).

kW is real power; kVA is apparent power including reactive power.

Yes, kVA is usually higher due to power factor losses.

A power factor of 0.9 or above is considered good.

kVA = kW / (√3 × PF).

Generators are rated in kVA because they supply both real and reactive power.

No, power factor ranges from 0 to 1.

Low power factor results in higher apparent power and inefficiency.

Yes, if the power factor and phase type are correctly applied.

Multiply the power in kilowatts by the time in hours. Formula: kWh = kW × hours.

kW is power (instant rate); kWh is energy (total used over time).

5 × 3 = 15 kWh.

Yes. Enter your panel’s power in kW and the hours of sun exposure.

Yes. You can estimate fuel usage by calculating energy output.

Break the period into intervals and sum each kW × hour segment.

Only if the device runs for 1 hour.

They charge based on kWh consumed.

10 × 0.15 = $1.50.

Yes. Divide kWh by time: kW = kWh ÷ hours.

Multiply kilowatts by 1000, then divide by the power factor. VA = kW × 1000 ÷ PF.

Most residential and commercial loads have a power factor between 0.8 and 1.0.

Yes, unless the power factor is 1, VA is always greater than or equal to kW.

kW is real power; VA is apparent power. VA includes both real and reactive power.

Yes, the conversion depends on the power factor of the load.

Yes, but ensure the total kW and power factor represent all phases.

Lower power factor increases VA, meaning larger equipment is needed.

Use an estimated value like 0.85 for general-purpose calculations.

Yes, generator sizing is based on VA, not just kW.

You may undersize or oversize your system, leading to inefficiency or failure.

Use the formula: V = (kW × 1000) ÷ (I × PF) for single-phase and V = (kW × 1000) ÷ (√3 × I × PF) for three-phase systems.

You need the power in kilowatts (kW), current in amps (A), power factor, and phase type (single or three-phase).

The power factor for most loads ranges from 0.8 to 1.0. Use 0.85 if you're unsure.

No, you need the current (amps) to calculate voltage from kW.

Yes, it supports both single-phase and three-phase voltage calculations.

√3 accounts for the phase shift between currents in a three-phase system, ensuring accurate voltage conversion.

Higher voltage reduces current and losses but must match the equipment rating.

Yes, it's helpful in calculating supply voltage when motor power and current are known.

It will affect accuracy. Use real measurements or a typical value (like 0.9).

No, this calculator is intended for AC systems (single-phase or three-phase).

Wh = kW × 1000 × Hours.

No, kW measures power, Wh measures energy over time.

1 kW for 1 hour equals 1000 Wh.

To measure energy consumption or production over time.

Yes, it's commonly used for sizing solar systems.

2.5 × 1000 × 3 = 7500 Wh.

Yes, it helps estimate battery capacity needed.

Yes, it's completely free and instant.

Input power in kilowatts and time in hours.

1000 Wh = 1 kWh, a common billing unit.

Divide the energy in kilowatt-hours (kWh) by time in hours: kW = kWh ÷ hours.

kW is power (rate), and kWh is energy (total usage over time).

Yes, it's useful for estimating solar panel output or load consumption.

Yes, it uses a standard engineering formula for energy-to-power conversion.

You need to input energy in kWh and time in hours.

Yes, use the reverse formula: kWh = kW × hours.

To determine how much power you used or plan capacity.

Yes, it's ideal for estimating power for appliances over time.

You must estimate or measure the duration of energy use.

Yes, the kWh to kW calculator is completely free.

Ohm's Law states that V = I × R.

A resistor is a component that resists the flow of current.

Voltage is the electric potential difference between two points.

Current is the flow of electric charge through a conductor.

Power is the rate at which electrical energy is consumed or produced.

The unit of resistance is Ohm, symbolized by Ω.

A capacitor is a device that stores electrical energy in an electric field.

An inductor is a component that stores energy in a magnetic field when current flows through it.

AC is alternating current; DC is direct current.

A transformer is a device that changes the voltage of an AC electrical supply.

A kilowatt-hour is a measure of energy equal to using 1,000 watts for 1 hour.

Use the formula: Watts = (kWh × 1000) ÷ Hours.

To find out the power consumption or rating of a device over time.

kWh measures energy, not power.

1 kWh equals 1,000 watt-hours.

You can't convert to watts without knowing the time duration.

Yes, it helps estimate appliance power use.

Watts measure power; kWh measures energy.

Yes, it helps you understand usage and cost breakdowns.

Yes, if the time and kWh are correctly entered.

Power (W) = Energy (J) ÷ Time (s).

Joules measure energy, watts measure power (energy per second).

To find power output or consumption from energy over time.

Energy in joules and time in seconds.

Yes, it helps calculate power usage of devices.

Yes, 1 watt = 1 joule/second.

Division by zero is undefined; time must be greater than zero.

Yes, the formula applies to both types of current.

Yes, multiply power (watts) by time (seconds) to get energy in joules.

Yes, if accurate energy and time values are used.

Amps = (kVA × 1000) ÷ Voltage for single-phase; for three-phase use Amps = (kVA × 1000) ÷ (√3 × Voltage).

Amps = (kVA × 1000) ÷ (√3 × Voltage).

No, power factor is not needed for kVA to amps conversion.

Yes, it provides reliable estimations for electrical load sizing.

No, frequency does not affect the kVA to amps calculation.

Yes, it's commonly used to estimate transformer load currents.

kVA stands for kilovolt-amperes, a unit of apparent power.

kW is real power; kVA includes reactive power and is apparent power.

Yes, kVA = (Amps × Voltage) ÷ 1000.

Yes, it helps calculate current for solar inverters and generators.

An eV is a unit of energy equal to 1.602×10⁻¹⁹ joules.

Use V = eV ÷ charge (in coulombs).

The elementary charge e = 1.602×10⁻¹⁹ coulombs.

Yes, when charge is 1 elementary charge, volts = eV ÷ e.

No, eV is a unit of energy, not voltage.

More charge reduces voltage for the same energy (V = E/Q).

It’s the energy gained by 1 electron across 1 volt.

In physics, electronics, quantum mechanics, and particle science.

Yes, it can indicate energy loss in physics contexts.

It measures energy levels in semiconductors and circuits.

Divide power (W) by current (A): V = P / I.

Power in watts (W), current in amperes (A), voltage in volts (V), resistance in ohms (Ω).

Yes — if you know resistance and power use V = √(P × R).

You can find current: I = P / V.

No. V = P / I. P = V × I is the correct relation for power.

Formulas assume DC or purely resistive AC loads; reactive AC needs RMS values and impedance handling.

Yes — convert to base units (W, A) first or use the unit selector if provided.

Sum total power first, then apply formulas with the combined power and appropriate current.

Yes — resistance can vary with temperature; use the resistance at operating temperature for best accuracy.

If a device uses 120 W at 2 A, voltage = 120 / 2 = 60 V.

Divide power (W) by current (A): V = P / I.

Use V = √(P × R) for resistive loads when power and resistance are known.

Use base units: watts (W), amperes (A), volts (V), ohms (Ω). Convert mW/mA/kW first.

Yes for purely resistive AC using RMS values; reactive AC needs impedance and phase considerations.

I = P / V.

Yes — P = V × I is the core relation for instantaneous power in DC and resistive circuits.

Sum individual powers for parallel loads, then use total power with total current to find system voltage.

Resistance can change with temperature; use operating-temperature resistance for best accuracy.

Yes — convert to amps/watts first or use a unit-aware input that converts automatically.

A 100 W load drawing 4 A requires V = 100 / 4 = 25 V.

Multiply duty cycle fraction by supply voltage: Vout = (Duty% / 100) × Vsup.

Duty cycle in %, voltage in volts (V).

Yes, a low-pass filter smooths PWM to a near-DC voltage close to Vout.

No, average voltage differs from RMS unless duty cycle and waveform match specific conditions.

Indirectly, by adjusting average voltage, PWM affects load current according to Ohm's law.

Recalculate Vout using the new supply voltage with same duty cycle.

Yes, use any value from 0 to 100% with decimal precision.

For motors with inertia, output speed correlates with average voltage, but torque and inductance affect response.

0 V — the output is always off.

Equal to the supply voltage — output is always on.

Multiply rated current (A) by coil resistance (Ω): V = I × R.

Yes, with a current-limited driver and within motor/driver limits.

It reduces inductive effects, improving speed and torque at high step rates.

An empirical method for estimating max driver voltage using coil inductance (L in mH).

Motor may overheat, insulation may fail, and driver could be damaged.

Yes, but torque and speed may be reduced.

Yes, chopper drivers allow higher voltage operation with current limiting.

Current in amperes (A), resistance in ohms (Ω), voltage in volts (V), inductance in millihenries (mH).

Yes, but use the correct coil resistance per wiring configuration.

Check datasheet or measure with an ohmmeter between motor phase wires.

Hall voltage is the voltage generated across a conductor in a magnetic field due to the Hall effect.

Use V_H = (B × I) / (n × q × t).

B in tesla, I in amperes, n in m⁻³, q in coulombs, t in meters, V_H in volts.

It’s used in magnetic field measurement, current sensing, and position detection.

Yes, via charge carrier density (n) and type of carriers.

Often in microvolts for metals, higher for semiconductors.

Yes, depending on the type of charge carriers (electrons or holes).

Yes, in wheel speed sensors, throttle position sensors, and current sensors.

Yes, by changing carrier density and mobility.

q is the elementary charge, about 1.602×10⁻¹⁹ coulombs.

Hall voltage is the voltage generated across a conductor in a magnetic field due to the Hall effect.

Use V_H = (B × I) / (n × q × t).

B in tesla, I in amperes, n in m⁻³, q in coulombs, t in meters, V_H in volts.

It’s used in magnetic field measurement, current sensing, and position detection.

Yes, via charge carrier density (n) and type of carriers.

Often in microvolts for metals, higher for semiconductors.

Yes, depending on the type of charge carriers (electrons or holes).

Yes, in wheel speed sensors, throttle position sensors, and current sensors.

Yes, by changing carrier density and mobility.

q is the elementary charge, about 1.602×10⁻¹⁹ coulombs.

It is the voltage measured between a live phase conductor and ground (earth).

Divide line-to-line voltage by √3.

Yes, in balanced systems.

It helps ensure safety, prevent overvoltage, and match equipment ratings.

Approximately 230V.

It may indicate a fault or incorrect grounding.

Only in a fault or open circuit condition.

V_LL is line-to-line voltage.

Because the phase voltage is the projection of line voltage in a balanced system.

Yes, especially in ungrounded or high-resistance grounded systems.

It calculates the reduced output voltage from a higher input voltage using a transformer or resistor divider.

Use a transformer and apply the turns ratio formula.

Use a resistor voltage divider or a regulator circuit.

Vout = Vin × (N2/N1).

Vout = Vin × (R2 / (R1 + R2)).

Yes, reducing voltage lowers shock and equipment damage risks.

Use a transformer with turns ratio of 12/230 ≈ 0.052.

No, transformers change voltage, not frequency.

Yes, by using transformers or DC-DC converters.

Yes, reversing it will step up voltage instead.

It calculates the transformer output voltage based on tap settings.

To match output voltage with supply and load conditions.

Vout = Vnominal × (1 + Tap%/100).

It increases output voltage by 5% over nominal.

It decreases output voltage by 5% below nominal.

Only with on-load tap changers (OLTC); off-load requires shutdown.

On-Load Tap Changer adjusts taps without interrupting power.

Typically within ±1% to ±2% of nominal voltage.

No, taps only change voltage level.

Yes, prolonged incorrect voltage can overheat or underpower loads.

The highest voltage a component or system can safely withstand under expected conditions.

Use V_peak = V_rms × √2.

Applying a safety factor (e.g., 80%) to a component’s rated voltage to extend life and reliability.

Compare component rating to peak (instantaneous) stress; convert AC RMS to peak first.

Depends on application; typical values range 0.6–0.95 — follow component datasheet recommendations.

Include transient margin in your comparison or use surge protection/clamping devices.

V_rated is the maximum specified — operating at or below a derated value is safer and increases lifetime.

DC ratings are steady-state; AC RMS must be converted to peak to check instantaneous stress.

Compare capacitor V_rated × derating factor to expected peak voltage; capacitors often need larger derating.

Lower temperatures can improve dielectric strength but always follow datasheet limits and derating guidance.

Use V = I × R, where I is steady current and R is coil resistance.

Compute Z = √(R²+(ωL)²) with ω=2πf, then V_rms = I_rms × Z.

Back-EMF is the induced voltage V_L = L·dI/dt generated when current changes in the coil.

Use V_L = L·dI/dt with worst-case dI/dt; include supply and clamp voltages for total stress.

Use volts (V), amperes (A), ohms (Ω), henries (H), hertz (Hz), and A/s for dI/dt.

Use V_rms from impedance at operating frequency and allow margin for back-EMF and driver headroom.

At AC or during transients, inductance increases impedance or causes induced voltage; steady DC sees only resistance.

Add flyback diodes (DC), snubbers, RC networks, TVS diodes or active clamp circuits.

Use P = I_rms² × R to estimate resistive (I²R) heating losses.

Yes — disconnect the coil and measure DC resistance with an ohmmeter; use low-current meters for low-resistance coils.

It converts milliamp current and resistance to voltage using Ohm’s Law.

V = (I_mA / 1000) × R.

To determine voltage drop or signal output in circuits.

Yes, divide mA by 1000 to get amps.

5 volts.

Yes, it’s ideal for current loop voltage calculations.

Yes, higher resistance produces higher voltage for the same current.

Yes, in AC or reverse current situations.

Yes, with RMS values for AC.

Only if resistance or voltage exceeds component ratings.

It is the ratio of total voltage to the product of the number of discs and voltage across the disc near the conductor, expressed in %.

Due to uneven voltage distribution caused by stray capacitance.

100% indicates perfect voltage sharing.

By using grading rings, capacitance grading, or longer strings.

It is directly connected to the line and sees the highest voltage drop.

Usually 70%–90% for HV systems without grading devices.

Yes, corona discharge can worsen uneven voltage distribution.

Disc spacing, string length, capacitance, and line voltage.

Mostly AC transmission lines, but also important in DC HV lines.

It is expressed as a percentage (%).

Use E = 4.44 f N Φ for induced RMS EMF per coil/phase.

Φ is the peak magnetic flux in webers (Wb) through one pole/coil.

Star: V_L = √3 × E; Delta: V_L = E.

Armature resistance and synchronous reactance cause drops: V_t ≈ E − I_a(R_a + jX_s).

f in Hz, Φ in Wb, N turns, E and V in volts (RMS).

4.44 ≈ 2π/√2, derived for sinusoidal flux converting peak to RMS for a single coil.

No — DC generator (commutator) requires different models; E = N dΦ/dt still applies for flux change.

Use turns per phase (total effective turns in series per phase) for N in the formula.

Yes — it gives open-circuit induced EMF; loaded behavior needs phasor analysis.

Compute complex V_t = E − I_a(R_a + jX_s) then take its magnitude |V_t| for RMS voltage value.

It is a tool to compute AC output voltage from DC input and efficiency.

Multiply the DC input voltage by the inverter efficiency.

It determines how much of the DC input is converted to usable AC output.

Yes, it is suitable for solar inverters and battery setups.

Voltage in volts, efficiency as a decimal or percentage.

No, due to energy losses, efficiency is always less than 100%.

Yes, extreme temperatures can reduce efficiency and voltage output.

Yes, but you may need to adjust for phase-specific calculations.

Yes, if input data is correct and efficiency is known.

You can estimate it from the inverter's specifications.

dBm is a power level expressed in decibels relative to 1 milliwatt.

Use P = 10^(PdBm/10) × 10^-3 watts.

Root Mean Square voltage represents the effective AC voltage.

Voltage is V_rms = sqrt(R × P) where R is resistance.

Impedance is measured in ohms (Ω).

Yes, provided you input the correct impedance value.

No, it applies to AC/RF signals and power measurements.

Impedance determines how voltage relates to power.

50 Ω is standard for RF systems.

Use the system’s characteristic impedance or measure it for accuracy.

dBm is a power level expressed in decibels relative to 1 milliwatt.

Use P = 10^(PdBm/10) × 10^-3 watts.

Root Mean Square voltage represents the effective AC voltage.

Voltage is V_rms = sqrt(R × P) where R is resistance.

Impedance is measured in ohms (Ω).

Yes, provided you input the correct impedance value.

No, it applies to AC/RF signals and power measurements.

Impedance determines how voltage relates to power.

50 Ω is standard for RF systems.

Use the system’s characteristic impedance or measure it for accuracy.

A sensor that measures temperature using the voltage generated by the Seebeck effect.

It generates EMF when two different metals are joined and exposed to a temperature difference.

The conversion of temperature difference directly into electric voltage.

Common types like K, J, T, E, N, R, S, and B.

Accuracy depends on input data and reference table precision.

Yes, to account for the reference junction temperature.

Yes, by using the NIST voltage-temperature conversion equations.

Yes, due to its wide temperature range and durability.

Millivolts (mV) for voltage and degrees Celsius (°C) for temperature.

No, if the wires are made from the correct thermocouple materials.

A tool to convert rotational speed into voltage using a voltage constant.

In motors, generators, alternators, and speed sensors.

It is the voltage constant, measured in volts per RPM.

From manufacturer data or by measuring voltage at a known RPM.

Yes, reversing rotation changes voltage polarity in DC generators.

Yes, but output depends on load and frequency.

RPM for speed and volts for output.

It is linear for many machines, but not all.

Yes, especially in generators without voltage regulation.

Yes, tachogenerators produce a voltage proportional to RPM.

The voltage measured when a source has no connected load.

In generators, alternators, transformers, and batteries.

Because there is no voltage drop due to load current.

Yes, directly proportional in AC machines.

Yes, with a voltmeter and no connected load.

Loss of residual magnetism or winding faults.

VNL = 4.44 × f × N × Φ.

Yes, but formula differs slightly.

Generally yes, but not for all machines.

Yes, as it changes winding resistance.

VA stands for volt-amps, a measure of apparent power.

Divide VA by current or use the square root of VA times resistance.

No, VA measures apparent power, watts measure real power.

VA accounts for both real and reactive power in AC systems.

Yes, to get watts from VA you multiply by the power factor.

Yes, but VA equals watts in DC systems.

VA for apparent power, volts for voltage, amps for current, ohms for resistance.

Yes, if you know the load resistance.

Yes, it determines transformer and UPS sizing.

Overload may occur, causing overheating or damage.

It’s a device or circuit that converts signal frequency into a proportional voltage.

Multiply frequency (Hz) by the sensitivity constant (V/Hz).

In tachometers, speed sensors, audio systems, and control circuits.

It’s the constant showing how many volts are produced per hertz.

Yes, depending on the F/V converter design.

Most circuits are linear, but some may have non-linear response curves.

Check the sensor datasheet or calibrate experimentally.

Yes, using code and an F/V conversion circuit.

Convert mV to V by dividing by 1000 before calculation.

Yes, filtering may be required for stable voltage output.

Use V = E / Q where E is joules and Q is coulombs.

Use V = sqrt(2E/C) where C is capacitance in farads.

Use V = Q / C.

Use E = 1/2 × C × V^2.

Energy in joules (J), charge in coulombs (C), capacitance in farads (F), voltage in volts (V).

Yes — convert units to farads first (µF → F = µF ×1e-6).

If Q=0 voltage is undefined; if E=0 then V=0 for any finite Q or C.

V=E/Q is a basic relation but batteries have internal resistance and inefficiencies to consider.

Because capacitor energy depends on V^2: E = 1/2 C V^2, so V ∝ sqrt(E/C).

You need very high voltage; check capacitor voltage rating and safety limits.

Use V = P / (I × PF) for single-phase or V = P / (√3 × I × PF) for three-phase AC.

Power factor is the ratio of real power to apparent power, between 0 and 1.

Because real power is lower for the same apparent power, requiring higher voltage for the same output.

No, PF is 1 in DC, so V = P / I.

Watts for P, amperes for I, volts for V, dimensionless PF.

Use a power meter that displays real power, apparent power, and calculates PF.

No, in practical systems PF ≤ 1.

Reactive power does not contribute to real work but affects PF and apparent power.

No, formula assumes balanced loads for three-phase systems.

Add power factor correction capacitors or synchronous condensers.

Multiply full-load current by service factor or by (1 + overload%/100).

The current drawn by a motor or device at its rated load and voltage.

A multiplier indicating how much over the rated load a motor can operate without damage.

Common values range from 1.0 to 1.25.

Set it slightly above full-load current, based on SF or manufacturer data.

The device may not trip in time, risking overheating and damage.

Nuisance tripping may occur during normal operation.

No, overload is sustained overcurrent; short-circuit is a fault current many times higher.

Yes, sustained overload causes overheating and insulation breakdown.

Thermal overload relays, circuit breakers, and fuses.

Divide voltage across insulation by insulation resistance.

Unwanted current flowing through insulation, moisture, or unintended paths.

Damaged insulation, moisture, contamination, or aging components.

Generally below 0.5 mA for handheld devices, per IEC/UL standards.

Using an insulation tester or leakage clamp meter.

IEC 60990 and UL 61010 provide guidelines.

Yes, if it exceeds the residual current device threshold.

Leakage current flowing to ground through insulation or chassis.

Yes, capacitive coupling can create AC leakage.

Improve insulation, keep equipment dry, and use RCD/GFCI protection.

Divide system voltage by total circuit impedance.

The current that flows during a short circuit or electrical fault.

It determines protective device ratings and system safety.

Low impedance in the fault path allows high current flow.

Fault current in a balanced 3-phase short circuit condition.

Fault current is measured in amperes (A).

Fault current assuming zero impedance at the fault location.

Yes, excessive current can overheat and destroy components.

IEEE 141 and IEC 60909 provide guidelines.

Increase system impedance or use current-limiting devices.

Multiply current (A) by voltage (V) and time (h) to get Wh, then divide by 1000 for kWh.

Energy is measured in watt-hours (Wh) or kilowatt-hours (kWh).

Yes, but for AC you may need to include the power factor.

Because 1 kWh equals 1000 Wh.

Time determines how long the current is flowing, affecting total energy.

E = I × V × t for Wh.

No, voltage is needed to calculate electrical energy.

It accounts for phase difference in AC systems and affects real energy.

Use average current or integrate over time for accuracy.

Yes, you can estimate battery energy based on current draw, voltage, and time.

It’s the ratio of actual output to theoretical output in an electrochemical process, expressed as a percentage.

Divide actual mass by theoretical mass and multiply by 100.

The mass predicted using Faraday’s law from the current and time applied.

Temperature, impurities, electrode material, and side reactions.

No, it’s usually less than or equal to 100%.

Higher efficiency means better material usage and lower costs.

Yes, it measures how much of the input current is stored as usable energy.

Optimize electrolyte composition, temperature, and current density.

A law that relates the amount of substance deposited to the electric charge passed.

It’s rare and requires ideal conditions with no losses.

It’s the unequal distribution of current between phases in a three-phase system.

Subtract the average current from the maximum current, divide by the average, then multiply by 100.

Typically less than 10% for most motors and equipment.

It causes overheating, efficiency loss, and can shorten motor life.

Uneven load distribution, faulty connections, or damaged conductors.

Balance loads, repair faulty wiring, or replace damaged components.

Yes, it can cause voltage imbalance which further stresses equipment.

During regular maintenance or if overheating or vibration is observed.

No, but they are related and often occur together.

Clamp meters, power quality analyzers, and motor testers.

It’s the flow of electric charge measured in amperes (A).

Use I=V/R if you know voltage and resistance, or I=P/V if you know power and voltage.

Current is measured in amperes (A).

Yes, it works for both AC and DC circuits.

It helps in selecting proper wire sizes, fuses, and ensuring safety.

You can use I=√(P/R) to find current.

Only if resistance stays the same.

Yes, I=V/R comes directly from Ohm’s Law.

Yes, in AC analysis, negative current indicates direction change.

No, voltage is potential difference, current is the flow of charge.

It’s the total amount of electrical current flowing through a circuit.

Use I=V/Rtotal, where V is voltage and Rtotal is the sum of all resistances.

Add all branch currents together.

Yes, in series it’s the same in all components, in parallel it splits and then sums.

Amperes (A).

Yes, in parallel circuits total current is the sum of branch currents.

Yes, in most cases they are the same.

Yes, higher voltage increases current if resistance stays the same.

Yes, it supports both AC and DC systems.

It ensures safe load limits and helps size electrical components.

Convert BTU/hr to kW using BTU/3412, then apply the amps formula.

I = (P*1000)/(V*PF).

I = (P*1000)/(√3*V*PF).

Yes, lower PF increases current draw.

Yes, but current may vary based on compressor speed.

Around 25A at 230V single-phase with PF 0.9.

About 34A at 230V single-phase, PF 0.9.

Possible causes include low voltage, dirty filters, or compressor issues.

Yes, enter total kW or BTU capacity.

It helps in choosing breakers, wiring, and avoiding overload.

It is the current drawn when a motor or equipment operates at its rated load.

I = (P*1000)/(V*PF).

I = (P*1000)/(√3*V*PF).

Yes, lower PF increases current draw.

Yes, convert HP to kW by multiplying by 0.746.

Possible causes include overload, low voltage, or mechanical issues.

Yes, if you know rated power, voltage, and power factor.

It ensures correct breaker, fuse, and wire sizing.

Yes, lower efficiency increases input power and current.

No, starting current is much higher than FLC.

It is the maximum instantaneous current value in a circuit.

Multiply RMS current by √2 for sine waves.

Yes, use I = Vpeak / R for resistive loads.

Yes, for AC sine waves it is √2 times higher.

Yes, in pulsed DC or transient events it is important.

Amperes (A) for current, volts (V) for voltage, ohms (Ω) for resistance.

Yes, excessive peak current can overheat or destroy components.

Surge current is a type of peak current occurring briefly at startup.

Use an oscilloscope or peak-hold ammeter.

It ensures components are rated to handle maximum stress.

Use I=2πfCVrms for sinusoidal AC, or I=Vrms/Xc with Xc=1/(2πfC).

Current leads voltage by 90° in a purely capacitive circuit.

C in farads, f in hertz, V in volts, current in amperes.

Use Ipeak=ωCVpeak with ω=2πf.

Use i(t)=C dv/dt or do a frequency-domain analysis with harmonics.

Higher frequency lowers Xc and increases current linearly with f.

Current is proportional to C; doubling C doubles current.

For steady DC, current is zero after the transient; use i=C dv/dt during changes.

Yes, ESR adds resistive heating and limits allowable ripple current.

Lower the voltage or frequency, or reduce capacitance.

Use I=Vrms/XL where XL=2πfL.

XL=2πfL, the opposition of an inductor to AC current.

Voltage leads current by 90° in a pure inductor.

Use i(t)=1/L ∫ v(t) dt + i(0) for general waveforms.

Higher frequency increases XL, lowering current.

In steady DC, an ideal inductor acts as a short circuit.

Yes, series resistance limits steady-state DC current.

τ=L/R controls rise or decay of current in transients.

Current cannot change instantly; it starts at its prior value.

Increase inductance, increase frequency, or reduce voltage.

The mean value of current over time, typically over one period for periodic waveforms.

Integrate i(t) over one period and divide by the period: Iavg=(1/T)∫0^T i(t) dt.

Over a full cycle it is zero.

Iavg = (2/π)·Ipeak.

Iavg = (1/π)·Ipeak.

Multiply duty cycle by the on-state current: Iavg = D·Ion.

No. RMS measures heating effect; average measures mean value.

For DC, Iavg = V/R. For PWM to a resistor, Ion ≈ V/R during on-time.

It indicates net DC delivery, battery charge/discharge, and converter outputs.

In pure AC (sine) with no DC offset.

It is the vector sum of the three phase currents; imbalance creates a non-zero current in the neutral conductor.

Use In = sqrt(Ia^2+Ib^2+Ic^2 - Ia*Ib - Ib*Ic - Ic*Ia).

Equal 120°-spaced currents cancel vectorially, resulting in ~0 A in the neutral.

Not directly; you input currents. Voltage is implicit if you derived Ia, Ib, Ic from load and LN voltage.

Set that phase current to 0 A; compute In from the remaining values.

With strong triplen harmonics it can, but at fundamental frequency it’s typically ≤ the largest phase current.

Yes. Triplen harmonics add arithmetically in the neutral and can raise In beyond the fundamental-only estimate.

No; delta has no neutral. Use this for 3-phase, 4-wire wye systems with LN loads.

It’s exact for fundamental-frequency phasors if phase angles are 120° and inputs are accurate RMS currents.

Balance single-phase loads across phases, limit non-linear loads per phase, and consider harmonic mitigation.

Use I = P / V (amps = watts ÷ volts).

I = P / (V · PF), where PF is 0–1 (use 1 for purely resistive loads).

Use I = P / (√3 · V_LL · PF) where V_LL is line-to-line voltage and PF is power factor.

For single-phase: I = S / V. For three-phase: I = S / (√3 · V_LL).

If PF < 1, current will be higher than P/V because real power is reduced by PF; use PF in the denominator.

If unknown and load is likely resistive, assume PF ≈1. For motors or inductive loads, assume PF 0.8–0.9 unless measured.

No. These formulas give steady-state RMS current from steady real power. Motor starting and inrush require separate transient calculations.

Choose next standard breaker rating above the continuous load plus apply code derating (e.g., 125% for continuous loads per many electrical codes). Verify with local rules.

No. Harmonics (from non-linear loads) increase RMS current; use measurements or harmonic analysis for accurate sizing.

Use line-to-line voltage (V_LL) with the √3 formula for line current. If you must use V_LN, convert or use phase formulas accordingly.

Id = Is (e^{Vd/(nVt)} - 1), where Vt = kT/q is the thermal voltage.

Is is the diode reverse saturation current — a tiny leakage current when the diode is reverse biased.

n (typically 1–2) models recombination and ideality; 1 for ideal diffusion-limited diodes, ~2 when recombination dominates.

Vt = kT/q. At 300 K, Vt ≈ 25.85 mV.

Before breakdown, reverse current ≈ −Is (small leakage). At breakdown voltage the simple model no longer applies.

At high forward currents, series resistance limits current; modify the model: Vd ≈ nVt ln(Id/Is+1) + Id Rs and solve numerically for Id.

Yes—at larger Vd the exponent can overflow. Use logarithmic math or limit Vd in calculators to avoid numeric overflow.

Multiply Id × Vd to get diode power dissipation (W) for thermal checks.

Shockley is a general model but parameters differ: LEDs/Schottkys have different Is, n, and Rs. Use device datasheet parameters for accuracy.

Use the diode datasheet, or fit measured I–V data (log-Id vs Vd) to extract Is and n via curve fitting.

Line current is the current in a supply conductor of a multi-phase system; it’s the current you measure in each line conductor.

For balanced 3-phase: I_L = P / (√3 · V_LL · PF). Use S for apparent power: I_L = S / (√3 · V_LL).

Wye (star): I_L = I_phase. Delta: I_L = √3 · I_phase.

Use line-to-line voltage V_LL for the standard three-phase power formulas. For wye phase calculations use V_LN accordingly.

Convert kW to W (×1000) then use I = P/(V·PF) for single-phase or the three-phase formula for 3Φ.

Yes for real power (kW). If you only have VA (apparent power), PF is not needed; use S instead of P.

Depends on connection: in delta line current = √3·I_phase so it can be larger; in wye they are equal.

It’s exact for balanced sinusoidal systems. For unbalanced loads, compute each phase or measure with clamp meters.

No—frequency doesn’t appear in the steady-state power/current algebraic formulas, but it affects impedance and transients in real systems.

Select a breaker rating above the continuous calculated current and apply code derating (e.g., 125% for continuous loads). Always follow local electrical codes.

Ripple current is the AC component of current that flows through a capacitor; it causes heating and limits capacitor life.

Use ΔVpp = I_load·Δt/C. For full-wave Δt=1/(2f); for half-wave Δt=1/f.

Rearrange: C = I_load·Δt / ΔVpp using the correct Δt for your topology.

I_c,rms quantifies heating in the capacitor from AC current. Use the capacitor’s RMS/ESR rating to avoid overheating.

For large-C capacitor input filters: full-wave I_c,rms≈0.483·I_load; half-wave I_c,rms≈0.707·I_load (engineering approximations).

Compute inductor ΔI_L, then I_c,rms≈ΔI_L/(2√3) (triangular approximation) and ΔVpp≈ΔI_L/(8 f_s C)+ESR·(ΔI_L/2).

Yes. ESR causes an instantaneous voltage drop from ripple current: V_ESR_peak ≈ I_c,pk · ESR; include it when checking ΔVpp.

Yes for voltage ripple and ripple current estimates, but check manufacturer RMS/ESR and temperature derating — ceramics have low ESR but may have capacitance change with voltage.

Select capacitors with adequate ripple-current rating, parallel multiple caps to share ripple, reduce ESR, or increase switching frequency to lower ΔI impact.

They are engineering approximations valid for common waveforms (short recharge pulses or triangular inductor ripple). For precise design use waveform integration, simulation, or measurements.

The small current drawn by a motor or transformer with no external mechanical or electrical load, mainly magnetizing plus loss components.

Use I0 = P0 / (√3 · V_LL · PF0), where P0 is measured no-load input power and PF0 is no-load power factor.

Use I0 = P0 / (V · PF0), with P0 the no-load input power, V the applied voltage and PF0 the no-load power factor.

Yes: I0 = Irated × (%I0 / 100) using the rated current and the percentage no-load value from the datasheet.

Transformers: ~1–5% of rated current; small induction motors: commonly 20–40%; large motors are usually lower.

Run the equipment unloaded at rated voltage and measure input power and line current with a clamp meter and power meter; compute I0 from P0 and PF0 if needed.

Yes—magnetizing current depends on flux (which follows voltage); overvoltage increases magnetizing and core-loss components, raising I0.

Yes—harmonics, especially from non-sinusoidal supply, can increase measured RMS no-load current and core heating beyond ideal calculations.

No—size breakers for expected load currents. Use no-load current only for idle-loss estimates and nuisance-trip checks; apply code derating for continuous loads.

Check for overvoltage, core saturation, damaged windings or incorrect connections; for motors ensure proper rotor/air-gap and for transformers consider core condition or using reactors/filters to limit magnetizing inrush or harmonics.

Norton current is the short-circuit current at the output terminals of a circuit as per Norton’s Theorem.

Short the load terminals, measure the resulting current—this is the Norton current (I_N).

Use I_N = V_TH / R_TH, where V_TH is Thevenin voltage and R_TH is Thevenin resistance.

Yes, for linear circuits, Norton current equals the short-circuit current across the load terminals.

Yes, Norton’s theorem applies to both AC and DC circuits, using impedance instead of resistance in AC.

They are dual forms—Norton’s ideal current source in parallel with R equals Thevenin’s voltage source in series with R.

It simplifies circuit analysis, making it easier to evaluate different loads and calculate currents.

Yes, the value depends on the source and network resistances or impedances in the original circuit.

Use appropriate current meters rated for the circuit’s voltage and expected short-circuit current.

Yes, if the network cannot deliver current to the shorted terminals, I_N will be zero.

Norton current is the short-circuit current at the output terminals of a circuit as per Norton’s Theorem.

Short the load terminals, measure the resulting current—this is the Norton current (I_N).

Use I_N = V_TH / R_TH, where V_TH is Thevenin voltage and R_TH is Thevenin resistance.

Yes, for linear circuits, Norton current equals the short-circuit current across the load terminals.

Yes, Norton’s theorem applies to both AC and DC circuits, using impedance instead of resistance in AC.

They are dual forms—Norton’s ideal current source in parallel with R equals Thevenin’s voltage source in series with R.

It simplifies circuit analysis, making it easier to evaluate different loads and calculate currents.

Yes, the value depends on the source and network resistances or impedances in the original circuit.

Use appropriate current meters rated for the circuit’s voltage and expected short-circuit current.

Yes, if the network cannot deliver current to the shorted terminals, I_N will be zero.

Line current is the current flowing through the external lines connecting the source to the load in a three-phase system.

Phase current is the current flowing through each individual phase winding or branch of a three-phase system.

In a star connection, line current equals phase current: I_line = I_phase.

In a delta connection, phase current = line current ÷ √3 and line current = phase current × √3.

Yes, it works for three-phase AC systems using either star or delta connections.

The √3 factor arises from the 120° phase difference between phases in a three-phase delta system.

No, only in star (Y) connections is line current equal to phase current. In delta, line current is √3 times phase current.

It is primarily for balanced three-phase systems; unbalanced loads require separate phase analysis.

It helps in sizing conductors, protective devices, and analyzing load distribution in three-phase circuits.

Compare with Ohm’s Law and the known voltages/impedances of the phases; line and phase currents should match the connection type.

It is the instantaneous current flowing into a capacitor as it charges in an RC circuit.

Use I(t) = (V / R) × e^(-t / (R × C)) where V is voltage, R resistance, C capacitance, and t time.

At t=0, the capacitor behaves like a short circuit, so I_initial = V / R.

The charging current decreases exponentially as the capacitor voltage approaches the supply voltage.

Time constant τ = R × C, it determines how fast the capacitor charges or discharges.

It is primarily for DC charging; AC requires impedance and reactance considerations.

The charging current will be small, and the capacitor will charge slowly.

A larger capacitance results in a longer charging time and a slower current decay.

Use V_C(t) = V × (1 - e^(-t / (R × C))) for the instantaneous voltage.

It helps in designing timing circuits, ensuring safe operation, and selecting appropriate resistors and capacitors.

It is converting electrical power in kilowatts to current in amperes using voltage and power factor.

Use I = KW × 1000 / (V × PF), where V is voltage and PF is power factor.

Use I = KW × 1000 / (√3 × V × PF) for balanced three-phase loads.

Use I = KW × 1000 / V, with V the DC voltage.

No, power factor only applies to AC circuits.

It helps in sizing conductors, protection devices, and determining load currents.

Yes, it supports both residential and industrial AC/DC circuits.

For AC, assume PF = 1 for an approximate calculation.

Yes, the formula works for balanced three-phase systems regardless of connection type.

Accuracy depends on correct voltage, power factor, and kilowatt input.

It is the value of current at a specific moment in time in a circuit.

Use i(t) = Imax × sin(ωt + φ), where Imax is peak current, ω is angular frequency, t is time, and φ is phase angle.

For steady-state DC, use i = V / R; for transients, use exponential functions.

It helps analyze waveforms, transients, and design safe, efficient circuits.

Amperes (A).

Instantaneous current is the value at one moment, while average current is over a full cycle or period.

Yes, with the correct time-domain equations.

Yes, in AC circuits it shifts the phase angle, affecting current timing.

Peak current is the maximum instantaneous value within a cycle.

Yes, it works for AC waveforms and DC transients or steady-state values.

It is the current flowing into the base terminal of a BJT to control the collector current.

Use Ib = Ic / β, where Ic is collector current and β is current gain.

They range from about 20 to 1000 depending on the transistor type.

It ensures the transistor is properly biased for amplification or switching.

Amperes (A).

Yes, higher β means less base current is needed for a given collector current.

No, it varies with temperature, biasing, and operating conditions.

No, without base current the BJT will not conduct in active mode.

hFE is the DC current gain of a BJT, the ratio of Ic to Ib.

No, MOSFETs use a gate voltage, not base current, for control.

For mechanical horsepower: kW = hp × 0.7457 (often rounded to 0.746).

Use hp = kW ÷ 0.7457 for mechanical horsepower.

Use kW = PS × 0.7355; conversely PS = kW ÷ 0.7355.

Yes, bhp is commonly treated as mechanical horsepower in conversions.

0.746 kW is a rounded engineering value; the precise factor is 0.7457 kW per hp.

The unit conversion does not; but sizing motors from mechanical load power requires dividing by efficiency.

Yes—automotive power in hp or PS can be converted to kW using the same constants.

No. kW is real power; kVA is apparent power. For AC systems kW = kVA × power factor.

Mechanical horsepower (hp/bhp). It also supports metric horsepower (PS).

They are exact by definition using 0.7457 (hp) and 0.7355 (PS); rounding may be applied for practicality.

kW = TR × 3.517, based on 1 TR = 12000 BTU/hr.

TR = kW ÷ 3.517.

It is the cooling capacity to freeze one ton of water in 24 hours.

Yes, it comes from 12000 BTU/hr × 0.00029307107 kW/BTU/hr.

TR is common in HVAC; kW is standard in electrical engineering.

Yes, most air conditioners are rated in TR or kW.

No, but actual electrical demand will be higher due to system losses.

5 × 3.517 = 17.585 kW.

50 ÷ 3.517 ≈ 14.22 TR.

Mostly in Asia; many other regions use kW or BTU/hr.

TR = kW ÷ 3.517.

kW = TR × 3.517.

It is the cooling capacity to freeze one ton of water in 24 hours.

Yes, from 12000 BTU/hr × 0.00029307107 kW/BTU/hr.

To match cooling capacity between electrical and HVAC standards.

Yes, it works for air conditioners, chillers, and refrigeration systems.

No, this is theoretical capacity; actual electrical use will be higher.

50 ÷ 3.517 ≈ 14.22 TR.

5 × 3.517 = 17.585 kW.

Mostly in Asia; many other regions use kW or BTU/hr.

Use Ah = C ÷ 3600 and mAh = C ÷ 3.6.

Use C = Ah × 3600.

Because 1 hour = 3600 seconds and 1 A = 1 C/s.

The charge delivered by 1 ampere flowing for 1 hour, equal to 3600 coulombs.

Use mAh for small batteries (phones, wearables) where capacities are fractional Ah.

No. Ah and C are charge units independent of voltage.

No. The unit conversion is fixed; temperature affects usable capacity, not the math.

Yes, coulomb counting integrates current over time to estimate state of charge.

Yes. 1 mAh × 3.6 = 3.6 C.

No. Wh = Ah × Voltage; Ah is charge, Wh is energy.

Multiply Ah by 3600 to get C.

Multiply mAh by 3.6 to get C.

Because 1 hour = 3600 seconds and 1 A = 1 C/s.

1 Ah equals 3600 C.

Ah measures charge, not energy.

No, Ah to C conversion is independent of voltage.

Yes, the unit conversion is universal.

Yes, exactly 3.6 coulombs.

Coulombs are the SI unit of electric charge.

Yes, divide C by 3600 to get Ah.

kW = bhp × 0.7457 (≈0.746).

bhp = kW ÷ 0.7457.

Yes, bhp is commonly treated as mechanical horsepower for conversions.

0.746 is a rounded engineering value; 0.7457 is more precise.

No—unit conversion is independent of efficiency, which affects required input power.

Yes, automotive bhp converts to kW using the same constant.

Metric hp uses 1 PS = 0.7355 kW; different from bhp.

No. kW is real power; kVA is apparent power (kW = kVA × PF).

Bhp typically refers to mechanical horsepower measured at the shaft; both use 0.7457 kW per hp.

Exact by definition using 0.7457; minor rounding may be used in practice.

W = BTU/h × 0.29307107.

BTU/h = W × 3.412141633.

It comes from the definition of the BTU and conversion from joules per second.

Yes, HVAC systems often list capacity in BTU/h which can be converted to watts.

Yes, the formula is the same for both cooling and heating capacity.

A BTU is the amount of heat required to raise 1 lb of water by 1°F.

Watts are the SI unit of power and used globally in electrical and mechanical engineering.

kW = (BTU/h × 0.29307107) ÷ 1000.

It’s about 293 watts—enough for a small space heater or cooling unit.

Yes, you can use 0.2931 W per BTU/h for quick estimates.

BTU/h = W × 3.412141633.

W = BTU/h × 0.29307107.

It comes from converting joules per second to BTUs per hour.

Yes, HVAC systems often rate capacity in BTU/h which can be converted to watts.

Yes, the formula is the same for both heating and cooling power.

A BTU is the heat required to raise 1 pound of water by 1°F.

Watts are the global SI unit of power, easier for scientific calculations.

kBTU/h = (W × 3.412141633) ÷ 1000.

Yes, exactly 1000 watts is 3412.141633 BTU/h.

Yes, 3.412 is often used for quick estimates.