JEE Physics Alternating Current Notes Set 03

AC and DC, Average and RMS Values

A current that reverses its direction periodically is alternating (AC); one that keeps a fixed direction is direct (DC). A sinusoidal current is written \( i=I_m\sin(\omega t+\phi) \), where \( \omega=2\pi f=\frac{2\pi}{T} \) is the angular frequency and \( (\omega t+\phi) \) is the phase.

The average of a function over a time interval is \( \langle f\rangle=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}f\,dt \), which is the net signed area under the graph divided by the time. A sinusoid averages to zero over a full cycle, and over a half cycle (starting from zero) it averages to \( \frac{2I_m}{\pi}\approx0.637I_m \).

The rms value is the square root of the mean of the square: \( I_{rms}=\sqrt{\langle i^2\rangle} \). It is the dc value that would produce the same heating. For a sine wave \( I_{rms}=\frac{I_m}{\sqrt2}\approx0.707I_m \) and \( V_{rms}=\frac{V_m}{\sqrt2} \). AC meters and the household figure of 220 V are rms values, so the peak is about 311 V.

For a dc part plus an ac part, \( I_{rms}^2=I_{dc}^2+\frac{I_m^2}{2} \). For \( i=a\sin\omega t+b\cos\omega t \), \( I_{rms}=\sqrt{\frac{a^2+b^2}{2}} \).

Power in an AC Circuit

For \( v=V_m\sin\omega t \) and \( i=I_m\sin(\omega t-\phi) \), the instantaneous power is \( vi \) and its average over a cycle is \( \langle P\rangle=V_{rms}I_{rms}\cos\phi \). The factor \( \cos\phi \) is the power factor. The component \( I_{rms}\sin\phi \) is the wattless current: it flows but does no net work.

The power oscillates at twice the supply frequency. In a resistor \( \phi=0 \) and \( P=I_{rms}^2R \); in a pure inductor or capacitor \( \phi=90^\circ \) and the average power is zero. Heat is produced only in the resistance.

Impedance is \( Z=\frac{V_m}{I_m}=\frac{V_{rms}}{I_{rms}} \). Inductive reactance is \( X_L=\omega L \) and capacitive reactance is \( X_C=\frac{1}{\omega C} \).

Circuits with a Single Element

Resistor: the current is in phase with the voltage, \( I_m=\frac{V_m}{R} \).

Capacitor: \( i=C\frac{dv}{dt}=\frac{V_m}{X_C}\cos\omega t \), so the current leads the voltage by \( 90^\circ \). \( X_C \) falls as the frequency rises and is infinite for steady dc, so a capacitor blocks dc.

Inductor: from \( v=L\frac{di}{dt} \) the current lags the voltage by \( 90^\circ \) and \( I_m=\frac{V_m}{\omega L} \). \( X_L \) grows with frequency and is zero for dc, so an inductor passes dc easily but chokes high frequencies.

Series RC, LR and LCR Circuits

In a series circuit the same current flows through every element, and the voltages add as phasors. For R and C, \( Z=\sqrt{R^2+X_C^2} \), \( \tan\phi=\frac{X_C}{R} \) and the current leads. For R and L, \( Z=\sqrt{R^2+X_L^2} \), \( \tan\phi=\frac{X_L}{R} \) and the current lags.

For the series LCR circuit: \( Z=\sqrt{R^2+(X_L-X_C)^2} \), \( \tan\phi=\frac{X_L-X_C}{R} \), and the source voltage is \( V=\sqrt{V_R^2+(V_L-V_C)^2} \). The voltages across L and C can individually exceed the source voltage. If \( X_L>X_C \) the circuit is inductive (current lags); if \( X_C>X_L \) it is capacitive (current leads).

Average power is \( P=I_{rms}^2R=V_{rms}I_{rms}\frac{R}{Z} \).

Resonance and Quality Factor

At the resonant angular frequency \( \omega_0=\frac{1}{\sqrt{LC}} \) (that is \( f_0=\frac{1}{2\pi\sqrt{LC}} \)) the reactances are equal, \( X_L=X_C \). Then \( Z=R \) is a minimum, the current \( \frac{V}{R} \) is a maximum, the power factor is 1 and the voltage across L equals that across C but they are opposite in phase.

The quality factor measures the sharpness of resonance: \( Q=\frac{\omega_0L}{R}=\frac{1}{\omega_0CR}=\frac1R\sqrt{\frac LC} \). A small R gives a large Q and a narrow, tall resonance curve. Radio tuning uses this: varying C changes \( f_0 \) to select a station.

At the half-power frequencies the net reactance equals R and the power is half the resonance value.

Parallel Circuits and Choke Coil

In a parallel circuit every branch has the same voltage and the branch currents add as phasors. For R, L and C in parallel, \( I=\sqrt{I_R^2+(I_C-I_L)^2} \) and \( \frac1Z=\sqrt{\frac1{R^2}+\left(\frac1{X_C}-\frac1{X_L}\right)^2} \).

A choke coil is a coil of large inductance and very small resistance used to cut down ac current. Unlike a series resistor it drops voltage without dissipating power, so it is the economical way to control ac.

Transformer

A transformer works by mutual induction between two coils on a common laminated soft-iron core, so it works only with changing current and not with steady dc. For an ideal transformer \( \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s} \). A step-up transformer has \( N_s>N_p \): it raises the voltage and lowers the current. Efficiency is \( \eta=\frac{V_sI_s}{V_pI_p} \).

Losses: copper loss (heating of the windings, reduced by thick wire), eddy-current loss (reduced by laminating the core), hysteresis loss (reduced by using soft iron) and flux leakage (reduced by winding the coils one over the other).

Worked Examples

Example 1. A coil carries 2.5 A from a 10 V battery, and 2 A from a 10 V, 50 Hz source. Find L. Solution: \( R=\frac{10}{2.5}=4\ \Omega \), \( Z=\frac{10}{2}=5\ \Omega \), so \( X_L=\sqrt{5^2-4^2}=3\ \Omega \) and \( L=\frac{3}{2\pi\times50}=\frac{3}{100\pi}\ \text{H} \).

Example 2. A 100 V, 100 W bulb must run at its rating from a 200 V, 50 Hz supply through a choke. Solution: \( R=\frac{100^2}{100}=100\ \Omega \) and the rated current is 1 A, so \( 1=\frac{200}{\sqrt{100^2+X_L^2}} \), giving \( X_L=100\sqrt3\ \Omega \) and \( L=\frac{\sqrt3}{\pi}\ \text{H} \).

Example 3. \( v=200\sqrt2\sin(\omega t+15^\circ) \) V drives a current \( i=2\sin(\omega t+45^\circ) \) A. The phase difference is \( 30^\circ \), so \( P=\frac{200\sqrt2}{\sqrt2}\cdot\frac{2}{\sqrt2}\cos30^\circ=100\sqrt6\ \text{W} \).