Alternating Current

JEE Main level, single correct, medium

171 of 190

Question 171 of 190

A transformer has an efficiency of 8080% and works at 10V10V and 4kW.4kW. If the secondary voltage is 240V,240V, then the current in the secondary coil is:

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Answer: (B) 13.33A13.33A

Efficiency η=PoutputPinput=ESISEPIP\eta =\dfrac{{P}_{output}}{{P}_{input}}=\dfrac{{E}_{S}{I}_{S}}{{E}_{P}{I}_{P}}
⇒0.8=240IS4000\Rightarrow 0.8=\dfrac{240{I}_{S}}{4000}
⇒IS=3200240=13.33A\Rightarrow {I}_{S}=\dfrac{3200}{240}=13.33A

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