PhysicsAlternating currentClass 12

AC circuits, phasors and resonance: interactive simulation

In an AC circuit the voltage across a resistor rises and falls with the current, across an inductor it peaks a quarter cycle earlier, and across a capacitor a quarter cycle later. Phasors turn these timing differences into arrows you can add.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards · 4 modes in one lab

What you can explore

  1. 1

    LCR & phasors

    R only, L only, C only and series LCR with rotating phasors and live v(t), i(t).
  2. 2

    Resonance & Q-factor

    Current against frequency for three resistances, with f₀, half-power points and bandwidth.
  3. 3

    Power factor

    Instantaneous power, true and apparent power, and the wattless component of current.
  4. 4

    Transformer

    Turns ratio, step-up and step-down, primary and secondary currents and efficiency.

Try this

Build an R, L, C or LCR circuit and watch the phasors spin, sweep the frequency through resonance, see why some current does no work at all, and step voltages up or down with a transformer.

  1. 1

    Series LCR with 220 V, 50 Hz, R = 100 Ω, L = 0.5 H, C = 40 μF.

    What you should see

    X_L ≈ 157 Ω, X_C ≈ 79.6 Ω, Z ≈ 126.5 Ω, I ≈ 1.74 A and the voltage leads by about 37.8°. Power ≈ 302 W.

  2. 2

    Switch to L only, then C only.

    What you should see

    V leads I by 90° for the inductor and lags by 90° for the capacitor; the power is zero in both.

  3. 3

    In Resonance, use L = 0.5 H, C = 20 μF, R = 20 Ω.

    What you should see

    f₀ ≈ 50.3 Hz, I_max = 220/20 = 11 A, Q ≈ 7.9 and bandwidth ≈ 6.4 Hz.

  4. 4

    Double R and then quadruple it.

    What you should see

    The peak stays at the same frequency but gets lower and wider — Q falls.

  5. 5

    In Transformer, use 200 and 1000 turns with 220 V and a 1000 Ω load.

    What you should see

    Step-up to 1100 V; the secondary carries 1.1 A and the primary 5.5 A, so the power is the same 1210 W on both sides.

Formulas behind it

Reactances

XL=ωL,XC=1ωCX_L = \omega L,\qquad X_C = \frac{1}{\omega C}

Impedance of series LCR

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Phase angle

tan⁡ϕ=XL−XCR\tan\phi = \frac{X_L - X_C}{R}

Resonant frequency

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

Quality factor

Q=ω0LR=1RLCQ = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

Average power

P=VrmsIrmscos⁡ϕP = V_{rms} I_{rms} \cos\phi

Ideal transformer

VsVp=NsNp=IpIs\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

Common mistakes

  • Adding V_R, V_L and V_C as plain numbers. They are 90° apart and must be added as phasors.
  • Saying the current is zero at resonance. It is maximum; only the net reactance is zero.
  • Thinking a step-up transformer increases power. It raises voltage and lowers current by the same factor.
  • Using peak values in P = VI cos φ. Both V and I must be rms values.
Practise 190 questionsCheck your answer instantly, with a worked solution.Stuck on a ac circuits, phasors and resonance question?Snap it and get a whiteboard video solution.

FAQ

What does wattless current mean?

It is the component I sin φ of the current that is 90° out of phase with the voltage. It moves energy back and forth but does no net work, so its average power is zero.

What is the power factor at resonance?

Exactly 1. X_L = X_C, so φ = 0 and the circuit behaves like a pure resistor.

Why is the voltage across L or C sometimes bigger than the supply voltage?

Near resonance V_L and V_C can each be Q times the supply, but they are opposite in phase and cancel, so their sum stays small.

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