JEE Physics Alternating Current Notes Set 05

Important Notes and Formulas

Alternating Current and its Averages

Direct current has a fixed direction and magnitude; alternating current changes both in step with time. For \( V=V_0\sin\omega t \) the average over a half cycle is \( V_{avg}=\frac1\pi\int_0^\pi V_0\sin\theta\,d\theta=\frac{2V_0}{\pi} \) (and similarly \( I_{avg}=\frac{2I_0}{\pi} \)); over a full cycle it is zero.
The root mean square value comes from \( V_{rms}^2=\frac1{2\pi}\int_0^{2\pi}V_0^2\sin^2\theta\,d\theta=\frac{V_0^2}{2} \), so \( V_{rms}=\frac{V_0}{\sqrt2} \) and \( I_{rms}=\frac{I_0}{\sqrt2} \).

AC Generator

An AC generator converts mechanical energy into electrical energy using electromagnetic induction. It has an armature coil, a field magnet, slip rings and brushes. For a coil of \( N \) turns and area \( A \) rotating at angular speed \( \omega \) in a field \( B \): \( e=NBA\omega\sin\omega t=e_0\sin\omega t \), with peak value \( e_0=NBA\omega \).

Transformer

A transformer changes the voltage of an alternating supply using mutual induction between a primary and a secondary coil on a common core. A step-up transformer has \( K>1 \) and increases the voltage while decreasing the current; a step-down transformer has \( K<1 \).
\( \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{i_p}{i_s}=K \) and efficiency \( \eta=\frac{\text{output power}}{\text{input power}}\times100\% \).

Example 1. A transformer has 100 turns in the primary and 1000 turns in the secondary. The primary is connected to a 220 V supply. Find the secondary voltage and the ratio of the currents (ideal transformer).
Solution:
\( V_s=V_p\frac{N_s}{N_p}=220\times10=2200 \) V, so it is a step-up transformer.
\( \frac{i_s}{i_p}=\frac{N_p}{N_s}=\frac1{10} \): the secondary current is one tenth of the primary current.

Pure Resistor, Capacitor and Inductor

Resistor: \( i=\frac{V_0}{R}\sin\omega t \); current is in phase with the voltage and the resistance does not depend on frequency.
Capacitor: \( q=CV \) gives \( i=C\frac{dV}{dt}=\omega CV_0\cos\omega t=\frac{V_0}{1/\omega C}\sin\left(\omega t+\frac\pi2\right) \). The current leads the voltage by \( \frac\pi2 \) and the capacitive reactance is \( X_C=\frac1{\omega C} \).
Inductor: \( V=L\frac{di}{dt} \) gives \( i=-\frac{V_0}{\omega L}\cos\omega t=\frac{V_0}{\omega L}\sin\left(\omega t-\frac\pi2\right) \). The current lags the voltage by \( \frac\pi2 \) and the inductive reactance is \( X_L=\omega L \).

Series LCR Circuit, Resonance and Power

The voltages add as phasors: \( V^2=V_R^2+(V_L-V_C)^2 \). The impedance is \( Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+\left(\omega L-\frac1{\omega C}\right)^2} \) and the phase angle satisfies \( \tan\phi=\frac{X_L-X_C}{R} \).
Electrical resonance occurs when \( X_L=X_C \), at \( f=\frac1{2\pi\sqrt{LC}} \); then \( Z=R \) is minimum and the current is maximum.
Average power: \( P=\frac12V_0i_0\cos\phi=V_{rms}i_{rms}\cos\phi \), where \( \cos\phi=\frac RZ \) is the power factor.

Example 2. In a series circuit \( R=30\ \Omega \), \( X_L=60\ \Omega \) and \( X_C=20\ \Omega \), connected to a 100 V (rms) supply. Find the impedance, the current, the power factor and the average power.
Solution:
\( Z=\sqrt{30^2+(60-20)^2}=50\ \Omega \) and \( I_{rms}=\frac{100}{50}=2 \) A.
\( \cos\phi=\frac RZ=0.6 \) and \( P=V_{rms}I_{rms}\cos\phi=100\times2\times0.6=120 \) W, which equals \( I^2R=4\times30 \).