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Q.

The secondary coil of an ideal step down transformer is delivering 500 watt power at 12.5 A current. If the ratio of turns in the primary to the secondary is 5: 1, then the current flowing in the primary coil will be :

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a

62.5 A

b

2.5 A

c

6 A

d

0.4 A

answer is B.

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Detailed Solution

P = VI
For secondary:
$V_{2} = \frac{P_{2}}{I_{2}} = \frac{500}{12.5} = 40 v o l t s$
For an ideal transformer (100% efficient)
$P_{i n p u t} = P_{o u t p u t} \Rightarrow V_{1} I_{1} = V_{2} I_{2}$
$\Rightarrow I_{1} = \frac{V_{2} I_{2}}{V_{1}} = \frac{40 (12.5)}{40 \times 5} = 2.5 A .$
$[∵ \frac{n_{1}}{n_{2}} = \frac{V_{1}}{V_{2}} \Rightarrow \frac{5}{1} = \frac{V_{1}}{40}]$

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