# NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

**1. A 100W resistor is connected to a 220V, 50 Hz AC supply.**

**(a) ****What is the rms v****alue 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 V_{0} = 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^{–3}H

V_{rms} = 220V

Frequency, f = 50 Hz

Inductive reactance X_{L} = 2*π*fL

= 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_{rms }cos 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^{–3}C

C = 30 µF (given in Q. 7)

= 30 × 10^{–6}F

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

f_{1} = 800 kHz = 8 × 10^{5} Hz

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

Maximum frequency f_{2 }= 1200 kHz = 12 × 10^{5} 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^{–12}F = 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^{–12}F = 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)

## NCERT Solutions for Class 12 Physics Chapter 7 Free PDF Download

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