(D) Given:Primary voltage,$V_p = 12\,kV = 12 \times 10^3\,V$Secondary voltage,$V_s = 120\,V$Power consumption,$P_s = 60\,kW = 60 \times 10^3\,W$For an

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(D) Given:Primary voltage,$V_p = 12\,kV = 12 \times 10^3\,V$Secondary voltage,$V_s = 120\,V$Power consumption,$P_s = 60\,kW = 60 \times 10^3\,W$For an ideal transformer,the power in the secondary circuit is given by $P_s = V_s \times I_s$.Therefore,the secondary current $I_s = \frac{P_s}{V_s} = \frac{60 \times 10^3}{120} = 500\,A$.Using Ohm's law for the secondary resistive load $R_s$:$R_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.24\,\Omega$.Converting to $m\Omega$:$R_s = 0.24 \times 10^3\,m\Omega = 240\,m\Omega$.

Class 12 Physics Electromagnetic Induction Transformer

Medium

An ideal transformer with a purely resistive load operates at $12\,kV$ on the primary side. It supplies electrical energy to a number of nearby houses at $120\,V$. The average rate of energy consumption in the houses served by the transformer is $60\,kW$. The value of the resistive load $(R_s)$ required in the secondary circuit will be $.........\,m\Omega$.

JEE MAIN 2023 All JEE MAIN

A
$250$
B
$260$
C
$290$
D
$240$

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The number of turns in the primary coil of a transformer is $200$ and the number of turns in the secondary coil is $10$. If $240 \ V$ $AC$ is applied to the primary,the output from the secondary will be......$V$.

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