Chapter 7 Alternating Current Class 12 Physics

Question 7.1

A $100 \Omega$ resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

(a) RMS Current

Given $R = 100 \Omega, V_{rms} = 220 \text{ V}$.

$$I_{rms} = \frac{V_{rms}}{R} = \frac{220}{100} = 2.2 \text{ A}$$

(b) Net Power Consumed

$$P = V_{rms} I_{rms} = 220 \times 2.2 = 484 \text{ W}$$

Result: (a) 2.2 A, (b) 484 W

Question 7.2

(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?

(a) RMS Voltage

Given $V_m = 300 \text{ V}$.

$$V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{300}{1.414} \approx 212.1 \text{ V}$$

(b) Peak Current

Given $I_{rms} = 10 \text{ A}$.

$$I_m = \sqrt{2} I_{rms} = 1.414 \times 10 = 14.14 \text{ A}$$

Result: (a) 212.1 V, (b) 14.14 A

Question 7.3

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

1. Inductive Reactance

$L = 44 \text{ mH} = 44 \times 10^{-3} \text{ H}$, $\nu = 50 \text{ Hz}$.

$$X_L = 2\pi\nu L = 2 \times 3.14 \times 50 \times 44 \times 10^{-3}$$ $$X_L = 100 \times 3.14 \times 0.044 = 13.82 \Omega$$

2. RMS Current

$$I_{rms} = \frac{V_{rms}}{X_L} = \frac{220}{13.82} \approx 15.92 \text{ A}$$

Result: 15.92 A

Question 7.4

A $60 \mu\text{F}$ capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

1. Capacitive Reactance

$C = 60 \mu\text{F} = 60 \times 10^{-6} \text{ F}$, $\nu = 60 \text{ Hz}$.

$$X_C = \frac{1}{2\pi\nu C} = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}}$$ $$X_C = \frac{1}{0.0226} \approx 44.2 \Omega$$

2. RMS Current

$$I_{rms} = \frac{V_{rms}}{X_C} = \frac{110}{44.2} \approx 2.49 \text{ A}$$

Result: 2.49 A

Question 7.5

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Explanation

Power in Inductive Circuit (Q7.3): In a pure inductor, current lags voltage by $90^\circ$ ($\phi = \pi/2$).
Power $P = V_{rms} I_{rms} \cos\phi = V_{rms} I_{rms} \cos(\pi/2) = 0$.

Power in Capacitive Circuit (Q7.4): In a pure capacitor, current leads voltage by $90^\circ$ ($\phi = -\pi/2$).
Power $P = V_{rms} I_{rms} \cos(-\pi/2) = 0$.

Net Power = Zero in both cases.

Question 7.6

A charged $30 \mu\text{F}$ capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Calculation

In an LC circuit, the angular frequency of free oscillations is given by:

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

$$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}$$ $$\omega = \frac{1}{\sqrt{810 \times 10^{-9}}} = \frac{1}{\sqrt{81 \times 10^{-8}}}$$ $$\omega = \frac{1}{9 \times 10^{-4}} = \frac{10000}{9} \approx 1.1 \times 10^3 \text{ rad/s}$$

Result: $1.1 \times 10^3 \text{ rad/s}$

Question 7.7

A series LCR circuit with $R = 20 \Omega$, $L = 1.5 \text{ H}$ and $C = 35 \mu\text{F}$ is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

1. Condition for Resonance

When supply frequency equals natural frequency, the circuit is in resonance. At resonance, $Z = R$.

2. Current at Resonance

$$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R} = \frac{200}{20} = 10 \text{ A}$$

3. Average Power

$$P = V_{rms} I_{rms} \cos\phi$$ $$\text{At resonance, } \phi = 0^\circ \implies \cos\phi = 1$$ $$P = 200 \times 10 \times 1 = 2000 \text{ W}$$

Result: 2000 W

Question 7.8

Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. $L = 5.0 \text{ H}$, $C = 80 \mu\text{F}$, $R = 40 \Omega$.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

(a) Resonant Frequency

$$\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}}$$ $$\omega_r = \frac{1}{20 \times 10^{-3}} = \frac{1000}{20} = 50 \text{ rad/s}$$ $$\nu_r = \frac{\omega_r}{2\pi} = \frac{50}{2\pi} \approx 7.96 \text{ Hz}$$

(b) Impedance and Current Amplitude

At resonance, $Z = R = 40 \Omega$.
RMS Current $I_{rms} = V_{rms}/R = 230/40 = 5.75 \text{ A}$.
Amplitude (Peak) Current $I_m = I_{rms}\sqrt{2} = 5.75 \times 1.414 \approx 8.13 \text{ A}$.

(c) Potential Drops

$$V_R = I_{rms} R = 5.75 \times 40 = 230 \text{ V}$$ $$V_L = I_{rms} X_L = I_{rms} (\omega_r L) = 5.75 \times (50 \times 5) = 1437.5 \text{ V}$$ $$V_C = I_{rms} X_C = I_{rms} \frac{1}{\omega_r C} = 5.75 \times \frac{1}{50 \times 80 \times 10^{-6}} = 1437.5 \text{ V}$$

Across LC Combination: $V_{LC} = V_L – V_C$ (since they are $180^\circ$ out of phase).
$V_{LC} = 1437.5 – 1437.5 = 0 \text{ V}$.

Result: (a) 50 rad/s, (b) 8.13 A, (c) $V_{LC} = 0$

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