Subject overview
Inductors
Inductance, magnetic energy, transformers, induced EMF, AC circuits and RL transients.
01 Inductors: Series & Parallel
Variables
- L1inductance 1
- Hhenry
- L2inductance 2
- Hhenry
- nnumber of inductors
- 1dimensionless
- Lparallel equivalent inductance
- Hhenry
Key Formulas
For a linear inductor, passive terminal voltage follows the current derivative; the induced opposing EMF has the opposite sign. Integrating power gives magnetic energy. A series RL step has time constant L/R.
Ideal positive inductances with no mutual magnetic coupling. Complete the visible inductances to add the next optional field. The empty final field is not included.
Calculate
Type prefixes directly: 1.5k, 250m or 20u. Results: 3 significant figures.
Ideal positive inductances with no mutual magnetic coupling. Complete the visible inductances to add the next optional field. The empty final field is not included.
Enter the required input values to calculate.
02 Transformer
Variables
- U1primary RMS voltage
- Vvolt
- U2secondary RMS voltage
- Vvolt
- I1primary RMS current
- Aampere
- I2secondary RMS current
- Aampere
- R1primary-equivalent resistance
- Ωohm
- R2secondary load resistance
- Ωohm
- P1input power
- Wwatt
- P2output power
- Wwatt
- N1primary turns
- 1dimensionless
- N2secondary turns
- 1dimensionless
- C1primary-equivalent capacitance
- Ffarad
- C2secondary capacitance
- Ffarad
- nturns ratio N₁/N₂
- 1dimensionless
- ηefficiency
- 1dimensionless
Key Formulas
A common time-varying core flux induces voltage in each winding in proportion to its turn count. Load reflection scales impedance by the square of the turns ratio. Lumped efficiency is an energy balance approximation.
Enter known quantities on either side; all uniquely determined missing quantities are calculated. Ideal voltage ratio with a resistive load and lumped efficiency 0 < η ≤ 1. If efficiency cannot be inferred, η = 1 is assumed explicitly. No magnetising current, saturation or winding voltage drop. Two values need not determine every quantity.
Calculate
Type prefixes directly: 1.5k, 250m or 20u. Results: 3 significant figures.
Enter known quantities on either side; all uniquely determined missing quantities are calculated. Ideal voltage ratio with a resistive load and lumped efficiency 0 < η ≤ 1. If efficiency cannot be inferred, η = 1 is assumed explicitly. No magnetising current, saturation or winding voltage drop. Two values need not determine every quantity.
Enter the required input values to calculate.
03 Inductor Calculations
Variables
- Icoil current
- Aampere
- Linductance
- Hhenry
- ALinductance factor
- Hhenry
- Nturns
- 1dimensionless
- μᵣrelative permeability
- 1dimensionless
- Across-sectional area
- m²square meter
- lmagnetic path length
- mmeter
- Φmagnetic flux per turn
- Wbweber
- Bflux density component
- Ttesla
- Wstored magnetic energy
- Jjoule
- ΔIcurrent change
- Aampere
- ΔΦflux change per turn
- Wbweber
- Δttime interval
- ssecond
- Uindmean induced coil EMF
- Vvolt
- lcmoving conductor length
- mmeter
- vconductor velocity component
- m/smeters per second
- Umotmotional EMF
- Vvolt
Key Formulas
For a linear inductor, passive terminal voltage follows the current derivative; the induced opposing EMF has the opposite sign. Integrating power gives magnetic energy. A series RL step has time constant L/R.
Linear ideal coil with constant permeability, uniform flux and no saturation, fringing or losses. Current I and flux Φ are distinct from their changes ΔI and ΔΦ. Coil EMF follows Lenz’s law; passive terminal voltage has the opposite sign. Motional EMF uses a separate conductor length and voltage, with conductor, velocity and field mutually perpendicular. Energy alone determines only the magnitude of current.
Calculate
Type prefixes directly: 1.5k, 250m or 20u. Results: 3 significant figures.
Linear ideal coil with constant permeability, uniform flux and no saturation, fringing or losses. Current I and flux Φ are distinct from their changes ΔI and ΔΦ. Coil EMF follows Lenz’s law; passive terminal voltage has the opposite sign. Motional EMF uses a separate conductor length and voltage, with conductor, velocity and field mutually perpendicular. Energy alone determines only the magnitude of current.
Enter the required input values to calculate.
04 RL at AC: Series & Parallel
Variables
- φphase angle
- radradian
- Itotal RMS current
- Aampere
- Rresistance
- Ωohm
- XLinductive reactance
- Ωohm
- Zimpedance magnitude
- Ωohm
- Linductance
- Hhenry
- Pactive power
- Wwatt
- QLinductive reactive power
- varvar
- Sapparent power
- VAvolt-ampere
- URresistive voltage
- Vvolt
- ULinductive voltage
- Vvolt
- Utotal RMS voltage
- Vvolt
- ffrequency
- Hzhertz
- cos(φ)power factor
- 1dimensionless
- sin(φ)reactive factor
- 1dimensionless
- IRresistive branch current
- Aampere
- ILinductive branch current
- Aampere
Key Formulas
For a linear inductor, passive terminal voltage follows the current derivative; the induced opposing EMF has the opposite sign. Integrating power gives magnetic energy. A series RL step has time constant L/R.
Switch between series and parallel RL connection. Series uses U_R and U_L; parallel uses I_R and I_L. U and I are RMS; φ, sin(φ) and Q_L are positive for an inductive load.
Calculate
Type prefixes directly: 1.5k, 250m or 20u. Results: 3 significant figures.
Ideal sinusoidal steady state. U and I are RMS; phase is voltage relative to supply current, negative for capacitive and positive for inductive loads. R, f and L > 0.
Enter the required input values to calculate.
Schematic curve; no numerical scale.
05 RL: Step Response & Discharging
Variables
- Isupply current
- Aampere
- i(t)instantaneous current
- Aampere
- τtime constant
- ssecond
- Rresistance
- Ωohm
- Linductance
- Hhenry
- Usupply voltage
- Vvolt
- u(t)instantaneous voltage
- Vvolt
- telapsed time
- ssecond
Key Formulas
For a linear inductor, passive terminal voltage follows the current derivative; the induced opposing EMF has the opposite sign. Integrating power gives magnetic energy. A series RL step has time constant L/R.
Switch between the RL step response and discharging current through R. Every listed quantity can be entered; uniquely determined missing values are calculated. R, L and τ > 0; t ≥ 0.
Calculate
Type prefixes directly: 1.5k, 250m or 20u. Results: 3 significant figures.
Ideal DC step U at t = 0, zero initial stored energy. R and L > 0; t ≥ 0.
Enter the required input values to calculate.
Schematic curve; no numerical scale.
Schematic curve; no numerical scale.