JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 28)
An ideal transformer with purely resistive load operates at $$12 ~\mathrm{kV}$$ on the primary side. It supplies electrical energy to a number of nearby houses at $$120 \mathrm{~V}$$. The average rate of energy consumption in the houses served by the transformer is 60 $$\mathrm{kW}$$. The value of resistive load $$(\mathrm{Rs})$$ required in the secondary circuit will be ___________ $$\mathrm{m} \Omega$$.
Answer
240
Explanation
The power delivered to the houses is given as 60 kW. This power is supplied at a voltage of 120 V. The power consumed in a resistive load can be found using the formula $P = V^2/R$, where P is the power, V is the voltage, and R is the resistance.
We can rearrange this formula to solve for the resistance:
$$R = V^2/P$$
Substituting the given values gives:
$$R = (120 \, \text{V})^2 / 60,000 \, \text{W} = 0.24 \, \Omega$$
Since we want the resistance in milliohms (mΩ), we can convert this to milliohms by multiplying by 1000:
$$R = 0.24 \, \Omega \times 1000 = 240 \, m\Omega$$
So, the value of resistive load required in the secondary circuit is 240 mΩ.
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