JEE Physics 2024 Question with Solution

Given: Primary voltage $V_p = 220 \, \text{V}$, primary turns $N_p = 100$, secondary turns $N_s = 10$. The secondary side is connected to two series resistors, and the output voltage $V_0$ is across the $7$ resistance.

Find: The value of $V_0$ and the correct option.

Use the transformer turns ratio first, then apply the voltage divider rule.

So the secondary voltage is $22 \, \text{V}$.

Now the two resistors are in series, so total resistance is

The output is across the $7$ resistance, so by voltage divider,

Therefore, the output voltage is $7 \, \text{V}$. The solution working gives $7 \, \text{V}$, which corresponds to option A. The solution says B, but that conflicts with the actual working and final value, so the defensible correct option is A.

Given: The transformer steps down $220 \, \text{V}$ to the secondary side using turns ratio $10:100$. The two resistors are in series with values $15$ and $7$.

Find: Output voltage across the $7$ resistor.

First compute the secondary voltage:

Then total series resistance is

So current in the secondary circuit is

Voltage across the $7$ resistor is

This shortcut works because once the total resistance becomes numerically equal to the secondary voltage, the current becomes $1$ in consistent units, making the drop across the resistor equal to its numerical resistance value. Hence the correct option is A.