Question

A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A the voltage across the secondary coil and the current in the primary coil respectively are

• A. 300 V, 15 A
• B. 450 V, 15 A
• C. 450 V, 13.5 A
• D. 600 V, 15 A

Hint:

For transformers, the power in the primary and secondary coils are related by efficiency and the voltage-current relationship.

Answer 1

The Correct Option is B

Given efficiency is 90%, and the power supply is 3 kW, so the output power of the secondary coil is: \[ P_{out} = 0.9 \times 3000 = 2700 \, \text{W} \] We can now use the formula for power: \[ P = V \times I \] From the given current in the secondary coil, $I_2 = 6$ A, so: \[ V_2 = \frac{P_{out}}{I_2} = \frac{2700}{6} = 450 \, \text{V} \] Now, using the power formula for the primary coil: \[ P_{in} = V_1 \times I_1 = 3000 \, \text{W} \] Using the efficiency, we find the primary current: \[ I_1 = \frac{P_{in}}{V_1} = \frac{3000}{200} = 15 \, \text{A} \]