NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

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1. A 100W resistor is connected to a 220V, 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?

Sol. Given, resistance R = 100 W, V_rms = 220 V Frequency, f = 50Hz

(a) Current flowing in the circuit

$$\text{I}_{rms}=\frac{\text{V}_{rms}}{\text{R}}=\frac{220}{100}=2.2\text{A}$$

(b) Net power consumed over a full cycle

P = V_rms × I_rms

= 220 × 2.2 = 484W

2. (a) The peak voltage of an AC supply is 300V. What is the rms voltage?

(b) The rms value of current in an AC circuit is 10A. What is the peak current?

Sol. (a) Given, peak value of AC voltage V₀ = 300V

The rms value of current I_rms = 10A

The rms value of voltage

$$\text{V}_{rms}=\frac{\text{V}_{0}}{\sqrt{2}}=\frac{300}{\sqrt{2}}=212.1\text{V}\\\text{(b)}\space \text{The rms value of current, I}_{rms} =\frac{\text{I}_{0}}{\sqrt{2}}\\\text{The peak value of current}\\\text{I}_{0}=\sqrt{2}\space\text{I}_{rms}\\=\sqrt{2}×10=14.14\text{A}$$

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

Sol. Given, inductance L = 44 mH = 44 × 10^–3H

V_rms = 220V

Frequency, f = 50 Hz

Inductive reactance X_L = 2πfL

= 2 × 3.14 × 50 × 44 × 10^–3

= 13.83

The rms value of current

$$\text{I}_{rms}=\frac{\text{V}_{rms}}{\text{X}_{L}}=\frac{220}{13.83}=15.9\text{A}$$

4. A 60µF capacitor is connected to a 110V, 60 Hz AC supply. Determine the rms value of the current in the circuit.

Sol. Given, capacitance C = 60µF = 60 × 10^–6 F

V_rms = 110V

Frequency of AC supply f = 60Hz

$$\text{Capacitive reactance X}_{C}=\frac{1}{2\pi fC}\\=\frac{1}{2×3.14×60×60×10^{\normalsize-6}}$$

= 44.23Ω

The rms value of the current

$$\text{V}_{rms}=\frac{V_{rms}}{X_{C}}=\frac{110}{44.23}=2.49\text{A}$$

5. In Q. 3 and 4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Sol. In Q. 3 Average power P = V_rms I_rmscos Φ

We know that the phase difference between current and voltage in case of inductor is 90°.

∴ P = V_rms I_rms cos 90° = 0

In Q. 4 Average power P = V_rms.I_rms cos Φ

We know that the phase difference between current and voltage in case of capacitor is 90° = 0

6. Obtain the resonant frequency ω of a series LCR circuit with L = 2.0 H, C = 32 µF and R = 10Ω. What is the Q-value of this circuit?

Sol. Resonant frequency

$$\omega_{r}=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{2×32×10^{\normalsize-6}}}\\=125\text{rad/s}\\\text{Q-value of this circuit,}\\\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{2}{32×10^{\normalsize-6}}}\\=\frac{10^{3}}{40}=25$$

7. A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Sol. Capacitance C = 30µF = 30 × 10^–6 F

Inductance L = 27 mH = 27 × 10^–3 H
For free oscillations, the angular frequency should be resonant frequency.

Resonant angular frequency

$$\omega_{r}=\frac{1}{\sqrt{LC}}\\=\frac{1}{\sqrt{27×10^{-3}×30×10^{-6}}}\\=\frac{10^{4}}{9}=1.1×10^{3}\text{rad/s}$$

8. Suppose the initial charge on the capacitor in Q. 7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Sol. Given, charge on the capacitor

Q = 6 mC = 6 × 10^–3C

C = 30 µF (given in Q. 7)

= 30 × 10^–6F

Energy stored in the circuit

$$\text{E}=\frac{\text{Q}^{2}}{2\text{C}}=\frac{(6×10^{-3})^{2}}{2×30×10^{\normalsize-6}}\\=\frac{30}{60}=0.6\text{J}$$

After some time, the energy is shared between C and L but the total energy remains same. So, we assume that there is no loss of energy.

9. A series LCR circuit with R = 20Ω, L = 1.5H and C = 35 µF is connected to a variable frequency 200 V AC supply. When the frequency of the supply equals the natural frquency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Sol. Given, resistance R = 20Ω, inductance L = 1.5H, capacitance C = 35µF = 35 × 10^–6 F and voltage V^rms = 200V.

When the frequency equal to the natural frequency of the circuit, this is the condition of resonance. At resonance, impedance Z = R = 20Ω

The rms value of current.

$$\text{I}_{rms}=\frac{\text{V}_{rms}}{\text{Z}}=\frac{200}{20}=10\text{A}\\\phi=0\degree\space(\text{for resonance})$$

Power transferred to the circuit over a complete cycle

P = I_rms.V_rms cos Φ

= 10 × 200 × cos 0°

= 2000W = 2kW

10. A radio can tune over the frequency range of a portion of MW broadcast band (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 mH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Sol. Given, minimum frequency

f₁ = 800 kHz = 8 × 10⁵ Hz

Inductance L = 200 µH = 200 × 10^–6H = 2 × 10^–4H

Maximum frequency f₂= 1200 kHz = 12 × 10⁵ Hz

For tuning, the natural frequency is equal to the frequency of oscillations that is the condition for resonance.

$$\text{Frequency of oscillations f =}\frac{1}{2\pi\sqrt{\text{LC}}}\\\text{For capacitance C}_1, \text{f}_1=\frac{1}{2\pi\sqrt{\text{LC}_{1}}}\\\text{C}_{1}=\frac{1}{4\pi^{2}f_{1}^{2}\text{L}}\\=\frac{1}{4×3.14×3.14×(8×10^{5})^{2}×2×10^{\normalsize-4}}$$

= 197.7 × 10^–12F = 197.7pF

$$\text{For capacitance C}_2, \text{f}_2 =\frac{1}{2\pi\sqrt{\text{LC}_{2}}}\\\text{C}_{2}=\frac{1}{4\pi^{2}f_{2}^{2}\text{L}}\\=\frac{1}{4×3.14×3.14×(12×10^{5})^{2}×2×10^{\normalsize-4}}$$

= 87.8 × 10^–12F = 87.8 pF

Thus, the range of capacitor is 87.8 pF to 197.7 pF.

11. Figure shows a series LCR circuit connected to a variable frequency 230V source.
L = 5.0 H, C = 80mF, R = 40W

(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.

Sol. (a) For resonance frequency of circuit

$$\omega_{r}=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{5×80×10^{\normalsize-6}}}$$

= 50 rad/s

Source frequency at resonance, then

$$v_{0}=\frac{\omega_{0}}{2\pi}=\frac{50}{2×3.14}=7.76\text{Hz}$$

(b) At the resonance, X_L = X_C

So, impedance of the circuit become

Z = R= 40Ω

The rms value of current

$$\text{I}_{rms}=\frac{\text{V}_{rms}}{\text{Z}}=\frac{230}{40}=5.75\text{A}\\\text{Amplitude of current I}_{0} = I_{rms}\sqrt2\\= 5.75 × \sqrt{2}= 8.13A\\\text{(c) The rms potential drop across L,}\\\text{V}_{L}=\text{I}_{rms}×X_{L}=\text{I}_{rms}×\omega_{r}\text{L}\\\text{= 5.75 × 50 × 5 = 1437.5V}\\\text{The rms potential drop across R}\\\text{V}_{R}=\text{I}_{rms}\text{R}=5.75×40=230\text{V}\\\text{The rms potential drop across C}_1,\\\text{V}_{C}=\text{I}_{rms}×X_{C}=\text{I}_{rms}×\frac{1}{\omega _{r} \text{C}}\\=5.75×\frac{1}{50×80×10^{\normalsize-6}}=1437.5\text{V}$$

Potential drop across LC combinations

= I_rms (X_L – X_C)

= I_rms (X_L – X_L) = 0

(∵ X_L = X_C at resonance)

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