## Alternating Current class 12 ncert solutions: Ncert Class 12 Physics Chapter 7 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 7 |

Chapter Name | Alternating Current class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 7 Alternating Current? Now you can download Ncert Class 12 Physics Chapter 7 Exercise Solutions pdf from here.

### Question 7.1: A 100 Ω 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?

**Solution 7.1:** Let’s break down the problem into its two parts.

**Part (a): RMS Value of Current**

**Given:**

- Resistance, R = 100 Ω
- Voltage, V
_{rms}= 220 V - Frequency, f = 50 Hz

The RMS (Root Mean Square) value of the current in the circuit can be calculated using Ohm’s Law:

\(I_{\text{rms}} = \frac{V_{\text{rms}}}{R}\)

Substituting the values:

\(I_{\text{rms}} = \frac{220 \, \text{V}}{100 \, \Omega} = 2.20 \, \text{A}\)

**Part (b): Net Power Consumed Over a Full Cycle**

The net power consumed in a resistive circuit is given by:

\(P = V_{\text{rms}} \times I_{\text{rms}}\)

Alternatively, since \(P = I_{\text{rms}}^2 \times R\), we can also calculate:

- \(P = (2.2 \, \text{A})^2 \times 100 \, \Omega \)
- \(= 4.84 \times 100 \, \text{W} \)
- =
**484 W**

**Conclusion:**

**(a)** The RMS value of the current in the circuit is **2.20 A**.

**(b)** The net power consumed over a full cycle is **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?

**Solution 7.2:**

**(a)**The peak voltage of the AC supply is V_{0}= 300 V.- We know that, V
_{rms} - \(V = \frac{ V _{0}}{ \sqrt{ 2 }}\)
- Substituting the values, we get \(V = \frac{ 300 }{ \sqrt{ 2 }}\)
- = 212.2 V
- Hence, the RMS voltage of the AC supply is 212.2 V.

**(b)**The RMS value of the current in the circuit is I = 10 A- We can calculate the peak current from the following equation,
- \(I _{ 0 } = \sqrt{ 2 } I\)
- \(10\sqrt{ 2 } = 14.1 A\)
- Therefore, the peak current is 14.1 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.

**Solution 7.3:** **Given:**

- Inductance, L = 44 mH = 44
**×**10^{-3}H - Voltage, V
_{rms}= 220 V - Frequency, f = 50 Hz

**Calculate the Inductive Reactance**

The inductive reactance ( X_{L} ) is given by: \(X_L = 2 \pi f L\)

Substitute the given values:

\(X_L = 2 \pi \times 50 \, \text{Hz} \times 44 \times 10^{-3} \, \text{H}\)

\(X_L = 2 \pi \times 50 \times 0.044\)

\(X_L \approx 2 \pi \times 2.2 \approx 13.82 \, \Omega\)

**Calculate the RMS Value of the Current**

Using Ohm’s Law for AC circuits:

\(I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L}\)

Substitute the values:

\(I_{\text{rms}} = \frac{220 \, \text{V}}{13.82 \, \Omega} \approx 15.9 \, \text{A}\)

The RMS value of the current in the circuit is approximately **15.9 A**.

### Question 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

**Solution 7.4:** Given:

- Capacitance, C = 60 μF = 60
**×**10^{-6}F - Voltage, V
_{rms}= 110 V - Frequency, f = 60 Hz

**Calculate the Capacitive Reactance**

The capacitive reactance ( X_{C }) is given by:

\(X_C = \frac{1}{2 \pi f C}\)

Substitute the given values:

\(X_C = \frac{1}{2 \pi \times 60 \, \text{Hz} \times 60 \times 10^{-6} \, \text{F}}\)

\(X_C = \frac{1}{2 \pi \times 60 \times 60 \times 10^{-6}}\)

\(X_C = \frac{1}{2 \pi \times 0.0036}\)

\(X_C \approx \frac{1}{0.0226} \approx 44.2 \, \Omega\)

**Calculate the RMS Value of the Current**

Using Ohm’s Law for AC circuits:

\(I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}\)

Substitute the values:

\(I_{\text{rms}} = \frac{110 \, \text{V}}{44.2 \, \Omega} \approx 2.49 \, \text{A}\)

The RMS value of the current in the circuit is approximately **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.

**Solution 7.5:** To determine the net power absorbed by each circuit, we need to consider the nature of each component and how it interacts with the AC supply:

1. **Inductive Circuit (from Exercise 7.3)**

**Component:**Inductor**RMS Voltage:**V_{rms}= 220 V**Inductance:**L = 44 mH**Inductive Reactance:**\( X_L \approx 13.82 \, \Omega \)**RMS Current:**\( I_{\text{rms}} \approx 15.9 \, \text{A} \)

**Power Absorbed by the Inductor:**

In a pure inductive circuit, the voltage and current are out of phase by 90 degrees. This phase difference means that the average power absorbed over a complete cycle is zero because the power factor \((\cos \phi)\) is zero (where \(\phi\) is the phase angle between voltage and current). The power absorbed by an inductor is purely reactive power, which does not contribute to net power dissipation.

**Net Power Absorbed:**P = 0 W

2. **Capacitive Circuit (from Exercise 7.4)**

**Component:**Capacitor**RMS Voltage:**V_{rms}= 110 V**Capacitance:**\( C = 60 \, \mu\text{F} \)**Capacitive Reactance:**\( X_C \approx 44.2 \, \Omega \)**RMS Current:**\( I_{\text{rms}} \approx 2.49 \, \text{A} \)

**Power Absorbed by the Capacitor:**

In a pure capacitive circuit, the voltage and current are also out of phase by 90 degrees. Like an inductor, the power factor \((\cos \phi)\) is zero in a purely capacitive circuit, meaning the average power absorbed over a complete cycle is zero. The capacitor also only stores and releases energy, but does not dissipate power as heat or work.

**Net Power Absorbed:**P = 0 W

**In both cases**, the net **power absorbed by each circuit** over a **complete cycle is zero**. This is because both the inductor and capacitor are reactive components that do not dissipate power but instead store and release energy.

### Question 7.6: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

**Solution 7.6:** Given capacitance value of the capacitor, C = 30 μF = 30 × 10^{–6} F

Given inductance value of the charged inductor, L = 27 mH = 27 × 10^{–3} H

Angular frequency is given as: \(\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }}\)

\(\omega _{ r } = \frac{ 1 }{ \sqrt{ 27 \times 10 ^{ – 3 } \times 30 \times 10 ^{ – 6 }}} \)

\(= \frac{ 1 }{ 9 \times 10 ^{ – 4 }} \)

\(= 1.11 \times 10 ^{ 3 }\, rad\,sec^{-1}\)

Therefore, the calculated angular frequency of free oscillation of the connection is 1.11 **x** 10^{3} s^{-1}.

### Question 7.7: A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF 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?

**Solution 7.7:** When the frequency of the AC supply equals the natural frequency of the series LCR circuit, the circuit is at resonance. At resonance, the reactances of the inductor and capacitor cancel each other out, and the circuit behaves as if it were purely resistive.

The natural (resonant) frequency ( f_{0} ) of a series LCR circuit is given by:

\(f_0 = \frac{1}{2 \pi \sqrt{LC}}\)

Given:

- L = 1.5 H
- C = 35 μF = 35
**×**10^{-6}F

First, calculate ( f_{0} ):

- \(f_0 = \frac{1}{2 \pi \sqrt{1.5 \times 35 \times 10^{-6}}}\)
- \(f_0 = \frac{1}{2 \pi \sqrt{52.5 \times 10^{-6}}}\)
- \(f_0 = \frac{1}{2 \pi \times 7.24 \times 10^{-3}}\)
- \( f_0 \approx 22.0 \, \text{Hz}\)

At resonance, the impedance ( Z ) of the circuit is equal to the resistance ( R ).

The impedance ( Z ) at resonance is:

Z = R = 20 Ω

The average power ( P ) transferred to the circuit in one complete cycle can be calculated using the formula:

\( P = \frac{V_{\text{rms}}^2}{R} \)

where V_{rms} is the root mean square (RMS) voltage of the AC supply. Given:

- V
_{rms}= 200 V - R = 20 Ω

Substitute these values into the formula:

\( P = \frac{(200)^2}{20}\)

\( P = \frac{40000}{20}\)

P = **2000 W**.

So, the average power transferred to the circuit in one complete cycle is **2000 W**.

### Question 7.8: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω

(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.

- Given that the inductance of the inductor in the circuit, L = 5.0 H
- Given that the capacitance of the capacitor in the circuit, C = 80 μF = 80 × 10
^{–6}F - Given that the resistance of the resistor in the circuit, R = 40 Ω
- Value of potential of the variable voltage supply, V = 230 V

**(a)**We know that the resonance angular frequency can be obtained by the following relation:- \(\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }}\)
- \(\omega _{ r } = \frac{ 1 }{ \sqrt{ 5 \times 80 \times 10 ^{ 6}}}\)
- \(\omega _{ r } = \frac{ 10 ^{ 3 }}{ 20 } = 50 rad/sec\)
- Thus, the circuit encounters resonance at a frequency of 50 rad/s.

**(b)**We know that the Impedance of the circuit can be calculated by the following relation:- \(Z = \sqrt{R^2 + \left(X_L – X_C\right)^2}\)
- At resonant condition,
- X
_{L}= X_{C} - Z = R = 40 Ω
- At resonating frequency, the amplitude of the current can be given by the following relation: \(I_{0} = \frac{V_{0}}{Z}\)
- Where, V
_{0 }= peak voltage = \(\sqrt{ 2 } V_{ rms }\) - Therefore,
- \(I_{0} = \frac{\sqrt{2} \, V_{\text{rms}}}{Z} \)
- \(= \frac{\sqrt{2} \times 230}{40} = 8.13 \, \text{A}\)
- Thus, at resonant condition, the impedance of the circuit is calculated to be 40 Ω, and the amplitude of the current is found to be 8.13 A

**(c) **The rms potential drop across the inductor in the circuit, ( V_{L} ) _{rms} = I x ω_{r} L

Where,

- \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{\sqrt{2} V}{\sqrt{2} Z} \)
- \(= \frac{230}{40} = \frac{23}{4} \, \text{A}\)
- Therefore, ( V
_{L})_{rms}\(\frac{ 23 }{ 4 } \times 50 \times 5 = 1437.5 V\)

We know that the potential drop across the capacitor can be calculated with the following relation:

- \(\left( V_{c} \right)_{\text{rms}} = I \times \frac{1}{\omega_{r} C} \)
- = \(\frac{ 23 }{ 4 } \times \frac{ 1 }{ 50 \times 80 \times 10 ^{ – 6 }} \)
- = 1437.5 V

We know that the potential drop across the resistor can be calculated with the following relation:

- \(\left( V_{R} \right)_{\text{rms}} = IR \)
- \(=\frac{ 23 }{ 4 } \times 40 \)
- = 230 V

Now, the potential drop across the LC connection can be obtained by the following relation: V_{LC} = I ( X_{L} − X_{C} )

At resonant condition,

X_{L} = X_{C}

V_{LC} = 0

Therefore, it has been proved from the above equation that the potential drop across the LC connection is equal to zero at a frequency at which resonance occurs.