JEE Physics Alternating Current Notes Set 01

Alternating Voltage and Current

A sinusoidal source gives \( V=V_0\sin\omega t \), with \( \omega=2\pi f \) (the mains frequency in India is 50 Hz). The mean value over a full cycle is zero. The mean over a half cycle is \( \frac{2V_0}{\pi}=0.637V_0 \).
The root-mean-square value is defined through heating: \( V_{rms}=\frac{V_0}{\sqrt2}=0.707V_0 \), \( I_{rms}=\frac{I_0}{\sqrt2} \). AC meters and the 220 V of the mains are rms readings; the peak of the mains is \( 220\sqrt2=311 \) V. For \( i=a+b\sin\omega t \), \( I_{rms}^2=a^2+\frac{b^2}{2} \). A hot-wire ammeter reads the rms value, with a non-uniform scale, and works for AC and DC.

Elements in an AC Circuit

Resistor: voltage and current in phase, \( I_0=\frac{V_0}{R} \).
Inductor: voltage leads current by 90 degrees, reactance \( X_L=\omega L=2\pi fL \) (zero for DC, rising with frequency).
Capacitor: current leads voltage by 90 degrees, reactance \( X_C=\frac1{\omega C} \) (infinite for DC, falling with frequency).
A phasor picture helps: take the current as reference; \( V_R \) is along it, \( V_L \) is 90 degrees ahead and \( V_C \) is 90 degrees behind.

Series LCR Circuit

Impedance \( Z=\sqrt{R^2+(X_L-X_C)^2} \) and source voltage \( V=\sqrt{V_R^2+(V_L-V_C)^2} \). Phase angle \( \tan\phi=\frac{X_L-X_C}{R} \): positive means inductive (current lags), negative means capacitive (current leads).
Series RL: \( Z=\sqrt{R^2+\omega^2L^2} \). Series RC: \( Z=\sqrt{R^2+\frac1{\omega^2C^2}} \).

Resonance

When \( X_L=X_C \) the circuit is in resonance at \( \omega_0=\frac1{\sqrt{LC}} \), \( f_0=\frac1{2\pi\sqrt{LC}} \). Then \( Z=R \) is minimum, the current \( \frac VR \) is maximum and in phase with the supply, and \( V_L \) and \( V_C \) cancel (each can be much larger than the supply).
Quality factor \( Q=\frac{\omega_0L}{R}=\frac1R\sqrt{\frac LC} \); the bandwidth is \( \Delta\omega=\frac RL \) in angular frequency. Smaller R gives a taller, sharper resonance curve, which is how radio tuning selects a station.

Power in AC Circuits

Average power \( P=V_{rms}I_{rms}\cos\phi \), where \( \cos\phi=\frac RZ \) is the power factor. Only the resistance dissipates energy: \( P=I_{rms}^2R \). In a pure inductor or capacitor \( \phi=90^\circ \) and the average power is zero (wattless current, component \( I_{rms}\sin\phi \)). At resonance \( \cos\phi=1 \).
A choke coil controls AC with its reactance without wasting power, unlike a series resistor.

Transformer

A step-up transformer has more turns in the secondary: \( \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s} \) for an ideal one. It works only with changing current. Losses: copper, eddy currents (reduced by laminating), hysteresis (reduced by soft iron) and flux leakage; there is no mechanical loss. Power is transmitted at high voltage to reduce the line loss \( I^2R \).

Solved Examples

Illustration 1. An alternating voltage V = 200 sin 100 pi t is applied across a 50 ohm resistor. Find the peak current, the rms current and the power dissipated.
Answer:
\( I_0=\frac{200}{50}=4 \) A, \( I_{rms}=\frac{4}{\sqrt2}=2.83 \) A.
\( P=I_{rms}^2R=8\times50=400 \) W.
4 A, 2.83 A, 400 W.

Illustration 2. A series circuit has R = 30 ohm, L = 0.1 H and C = 100 microfarad on a 100 V, 50 Hz supply. Find the impedance and the current.
Answer:
\( X_L=2\pi\times50\times0.1=31.4 \) Ω, \( X_C=\frac1{2\pi\times50\times10^{-4}}=31.8 \) Ω.
\( Z=\sqrt{30^2+(31.4-31.8)^2}\approx30 \) Ω, so \( I=\frac{100}{30}=3.33 \) A.
Z about 30 ohm, I about 3.3 A (near resonance).

Illustration 3. Find the resonant frequency and the Q factor for L = 2 mH, C = 20 microfarad and R = 5 ohm.
Answer:
\( f_0=\frac1{2\pi\sqrt{2\times10^{-3}\times20\times10^{-6}}}=\frac1{2\pi\times6.32\times10^{-5}}=796 \) Hz.
\( Q=\frac1R\sqrt{\frac LC}=\frac15\sqrt{100}=2 \).
f0 about 796 Hz, Q = 2.

Illustration 4. A coil of resistance 8 ohm and inductance 0.06 H is connected to a 100 V, 50 Hz supply. Find the current, the power factor and the power.
Answer:
\( X_L=2\pi\times50\times0.06=18.85 \) Ω, \( Z=\sqrt{64+355.3}=20.5 \) Ω.
\( I=\frac{100}{20.5}=4.88 \) A, \( \cos\phi=\frac{8}{20.5}=0.39 \).
\( P=VI\cos\phi=100\times4.88\times0.39=190 \) W.
I about 4.9 A, power factor about 0.39, P about 190 W.

Illustration 5. A step-up transformer has 100 primary turns and 2000 secondary turns. The primary is at 220 V and draws 10 A. Find the secondary voltage and current for an ideal transformer.
Answer:
\( V_s=220\times20=4400 \) V.
\( I_s=\frac{I_pN_p}{N_s}=\frac{10}{20}=0.5 \) A.
4400 V and 0.5 A.