Q117 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
An ideal transformer converts $220$V a.c. to $3.3$ KV a.c. to transmit a power of $4.4$ KW. If primary coil has $600$ turns then alternating current in the secondary coil is
- Correct answerA$\dfrac{4}{3}$ A
- B$\dfrac{3}{4}$ A
- C$\dfrac{1}{3}$ A
- D$\dfrac{2}{3}$ A
Solution
For an ideal transformer, the input power is equal to the output power.
$P_{out} = P_{in} = 4.4 \text{ kW} = 4400 \text{ W}$
The output power is given by $P_{out} = V_s I_s$
$4400 = 3300 \times I_s$
$I_s = \dfrac{4400}{3300} = \dfrac{4}{3} \text{ A}$
Answer: $\dfrac{4}{3}$ A
$P_{out} = P_{in} = 4.4 \text{ kW} = 4400 \text{ W}$
The output power is given by $P_{out} = V_s I_s$
$4400 = 3300 \times I_s$
$I_s = \dfrac{4400}{3300} = \dfrac{4}{3} \text{ A}$
Answer: $\dfrac{4}{3}$ A
Hence, the correct answer is option: 'A'.