This set of Class 12 Physics Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Alternating Current”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Which among the following is true about transformers?

a) Transformers are used to convert low alternating voltage to a high alternating voltage

b) Transformers are used to convert low alternating current to a high alternating current

c) Transformers are based on the phenomena of mutual electric field

d) Transformers are used only for low alternating voltage

View Answer

Explanation: Transformers are used to convert low alternating voltage to a high alternating voltage and vice versa. Transformers are based on the phenomena of mutual induction. A transformer consists of a soft iron coil with two coils wound around it which are not connected to one another. The other statements are not valid.

2. Identify the expression for transformation ratio from the following.

a) k = \(\frac {V_s}{V_p}=\frac {I_s}{I_p}=\frac {N_s}{N_p}\)

b) k = \(\frac {V_s}{V_p}=\frac {I_s}{I_p}=\frac {N_p}{N_s}\)

c) k = \(\frac {V_s}{V_p}=\frac {I_p}{I_s}=\frac {N_s}{N_p}\)

d) k = \(\frac {V_p}{V_s}=\frac {I_s}{I_p}=\frac {N_s}{N_p}\)

View Answer

Explanation: Transformation ratio (k) is defined as the ratio of number of turns in the secondary winding to the number of turns in the primary winding and the ratio of the current in the primary coil to the current in the secondary coil. It also defined as the ratio of voltage at the secondary terminal to the voltage at the primary terminals. The expression is given as:

k = \(\frac {V_s}{V_p}=\frac {I_p}{I_s}=\frac {N_s}{N_p}\)

3. Pick out the correct combination for a step-up transformer.

a) k < 1; V_{s} > V_{p}, I_{s} > I_{p}, N_{s} > N_{p}

b) k > 1; V_{s} > V_{p}, I_{s} < I_{p}, N_{s} > N_{p}

c) k > 1; V_{s} > V_{p}, I_{s} > I_{p}, N_{s} > N_{p}

d) k < 1; V_{s} < V_{p}, I_{s} > I_{p}, N_{s} > N_{p}

View Answer

Explanation: For a step-up transformer, the voltage at the second terminal is greater than the primary terminal, i.e. ➔ V

_{s}> V

_{p}. Similarly, the current in the secondary coil is lesser than that current flowing the primary coil, i.e. ➔ I

_{s}< I

_{p}. Also, the number of turns in the secondary winding is greater than that in the primary winding, i.e. ➔ N

_{s}> N

_{p}. So the transformation ratio will be greater than 1. So, the correct combination is given as:

**k > 1; V**

_{s}> V_{p}, I_{s}< I_{p}, N_{s}> N_{p}4. The efficiency of a transformer is the ratio of output power to input power.

a) True

b) False

View Answer

Explanation: Yes, the efficiency of a transformer can be determined by taking the ratio of output power to input power. Transformers are the most highly efficient electrical devices. Most of the transformers have full load efficiency between 95% and 98.5%.

Efficiency (η)=\(\frac {Output \, power}{Input \, power}=\frac {V_s I_s}{V_p I_p}\)

5. Identify the correct combination for a step-down transformer.

a) k < 1; V_{s} > V_{p}, I_{s} > I_{p}, N_{s} < N_{p}

b) k > 1; V_{s} < V_{p}, I_{s} > I_{p}, N_{s} > N_{p}

c) k > 1; V_{s} > V_{p}, I_{s} < I_{p}, N_{s} > N_{p}

d) k < 1; V_{s} < V_{p}, I_{s} > I_{p}, N_{s} < N_{p}

View Answer

Explanation: For a step-down transformer, the voltage at the second terminal is lesser than the primary terminal, i.e. ➔ V

_{s}< V

_{p}. Similarly, the number of turns in the secondary winding is lesser than that in the primary winding, i.e. ➔ N

_{s}< N

_{p}. Also, the current in the secondary coil is greater than that current flowing the primary coil, i.e. ➔ I

_{s}> I

_{p}. So the transformation ratio will be lesser than 1. So, the correct combination is given as:

**k < 1; V**

_{s}< V_{p}, I_{s}> I_{p}, N_{s}< N_{p}6. Which of the following is usually taken to make the core of a transformer?

a) Aluminum

b) Copper

c) Soft iron

d) Hard iron

View Answer

Explanation: Core of a transformer is used to transfer the flux produced by primary coils of transformer to secondary coils of a transformer. Therefore, the core of a transformer is made of soft iron because it has high permeability so it provides complete linkage of magnetic flux of the primary coil to the secondary coil.

7. A transformer has an efficiency of 60% and works at 5kW. If the secondary voltage is 150 V, then what is the secondary current?

a) 10 A

b) 20 A

c) 30 A

d) 40 A

View Answer

Explanation: Given: efficiency (η) = 60%; Input power (P

_{i}) = 5 kW = 5000 W; V

_{S}= 150 V

Efficiency (η)=\(\frac {Output \, power (P_o)}{Input \, power (P_i)}=\frac {V_s I_s}{V_p I_p}\)

\(\frac {60}{100}=\frac {P_o}{5000}\)

P

_{o}=\(\frac {60 \times 5000}{100}\)=3000 W

Thus,

I

_{s}=\(\frac {P_o}{V_s} = \frac {3000}{150}\)=20 A

Therefore, the current in the secondary coil is 20 A.

8. Current remains unchanged in a transformer.

a) True

b) False

View Answer

Explanation: No, this statement is false. The quantity that remains unchanged in a transformer is the frequency of the input alternating current. Frequency of the alternating current at input = Frequency of the alternating current at output.

9. A transformer is used to light 100 W 25 volt lamp from 250 Volt ac mains. The current in the main cable is 0.5 A. Calculate the efficiency of the transformer.

a) 50%

b) 60%

c) 90%

d) 80%

View Answer

Explanation: Given: Output power (P

_{o}) = 100 W; I

_{p}= 0.5 A; V

_{p}= 250 V

Input power (P

_{i}) = V

_{p}× I

_{p}

P

_{i}= 250 × 0.5 = 125 W

Efficiency = \((\frac {Output \, power}{Input \, power})\) × 100%

Efficiency = \((\frac {100}{125})\) × 100

Efficiency = 80

Therefore, the efficiency of the transformer is 80%.

10. In a step-down transformer, the number of turns in the secondary coil is 20 and the number of turns in the primary coil is 100. If the voltage applied to the primary coil is 120 V, then what is the voltage output from the secondary coil?

a) 24 V

b) 12 V

c) 6 V

d) 18 V

View Answer

Explanation: Given: N

_{s}= 20; N

_{p}= 100; V

_{p}= 120 V

For a step-down transformer ➔ \(\frac {N_s}{N_p} = \frac {V_s}{V_p}\)

V

_{s}=\(\frac {N_s \times V_P}{N_p}\)

V

_{s}=\(\frac {20 \times 120}{100}\)

**V**

_{s}=24 VTherefore, the voltage output from the secondary coil is 24 V.

**More MCQs on Class 12 Physics Chapter 7:**

- Chapter 7 – Alternating Current MCQ (Set 2)
- Chapter 7 – Alternating Current MCQ (Set 3)
- Chapter 7 – Alternating Current MCQ (Set 4)
- Chapter 7 – Alternating Current MCQ (Set 5)
- Chapter 7 – Alternating Current MCQ (Set 6)
- Chapter 7 – Alternating Current MCQ (Set 7)
- Chapter 7 – Alternating Current MCQ (Set 8)
- Chapter 7 – Alternating Current MCQ (Set 9)
- Chapter 7 – Alternating Current MCQ (Set 10)

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