A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is

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Solution

Q.
A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is [2024]

1 1.59A

2 13.33 A

3 1.33 A

4 15.1 A

Ans.
(2)

$Efficiency η = 80 %, P_{i} = 4 kW = 4000 W$

   $V_{P} = 10 V,$ $V_{S} = 240 V$

   $For primary coil : I_{P} = \frac{P_{i}}{V_{P}} = \frac{4000}{10} = 400 A$

   $η = \frac{V_{S} I_{S}}{V_{P} I_{P}} \Rightarrow \frac{80}{100} = \frac{240 I_{S}}{10 \times 400}$

$\Rightarrow I_{S} = 13.33 A$

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