Alternating Current — NCERT Solutions

In a pure inductor and a pure capacitor, the current and voltage are out of phase by $\pi/2$. The average power over a complete cycle is zero:

• For an inductor, the text shows the average power over one cycle is zero.
• For a capacitor, the text also shows the average power over one cycle is zero.

So in Exercises 7.3 and 7.4, the net power absorbed by each circuit over a complete cycle is $0\,\text{W}$.

For free oscillations of an LC circuit,

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

Given

$$L = 27\,\text{mH} = 27\times 10^{-3}\,\text{H}, \quad C = 30\,\mu\text{F} = 30\times 10^{-6}\,\text{F}$$

So,

$$\omega_0 = \frac{1}{\sqrt{(27\times 10^{-3})(30\times 10^{-6})}}$$

$$= \frac{1}{\sqrt{8.1\times 10^{-7}}}$$

$$\approx \frac{1}{9.0\times 10^{-4}} \approx 1.11\times 10^3\,\text{rad s}^{-1}$$

Thus, the angular frequency is $1.11\times 10^3\,\text{rad s}^{-1}$.

At resonance in a series LCR circuit, the impedance becomes

$$Z = R$$

and the phase angle is zero, so power factor is 1. Hence the average power is

$$P = \frac{V^2}{R}$$

Given $V = 200\,\text{V}$ and $R = 20\,\Omega$,

$$P = \frac{(200)^2}{20} = \frac{40000}{20} = 2000\,\text{W}$$

So the average power transferred is $2000\,\text{W}$.

Given a series LCR circuit with

$$L = 5.0\,\text{H},\quad C = 80\,\mu\text{F} = 80\times 10^{-6}\,\text{F},\quad R = 40\,\Omega$$

and source voltage $V = 230\,\text{V}$.

(a) Resonant frequency

For resonance,

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

So,

$$\omega_0 = \frac{1}{\sqrt{5.0\times 80\times 10^{-6}}}$$

$$= \frac{1}{\sqrt{4.0\times 10^{-4}}} = \frac{1}{0.02} = 50\,\text{rad s}^{-1}$$

Frequency:

$$\nu_0 = \frac{\omega_0}{2\pi} = \frac{50}{2\pi} \approx 7.96\,\text{Hz}$$

(b) Impedance and current at resonance

At resonance,

$$Z = R = 40\,\Omega$$

Rms current:

$$I = \frac{V}{Z} = \frac{230}{40} = 5.75\,\text{A}$$

(c) RMS potential drops across the elements

At resonance,

$$X_L = X_C$$

Calculate them:

$$X_L = \omega_0 L = 50\times 5.0 = 250\,\Omega$$

$$X_C = \frac{1}{\omega_0 C} = \frac{1}{50\times 80\times 10^{-6}} = 250\,\Omega$$

Therefore,

$$V_R = IR = (5.75)(40) = 230\,\text{V}$$

$$V_L = I X_L = (5.75)(250) = 1437.5\,\text{V}$$

$$V_C = I X_C = (5.75)(250) = 1437.5\,\text{V}$$

Since $V_L$ and $V_C$ are equal in magnitude and opposite in phase, their vector sum is zero:

$$V_L + V_C = 0$$

So the potential drop across the LC combination is zero at resonance. The entire source voltage appears across the resistor.