JEE Physics 2023 Question with Solution

Given: Thermal energy developed is H for current $4 \, \text{A}$ in $10 \, \text{s}$.

Find: Thermal energy developed when current becomes $16 \, \text{A}$ for the same time.

For a resistor, thermal energy is proportional to the square of current when time is constant.

$$H \propto i^2$$

Therefore,

$$\frac{H}{H'} = \left(\frac{4}{16}\right)^2$$ $$\frac{H}{H'} = \frac{1}{16}$$ $$H' = 16H$$

Therefore, the thermal energy developed is 16H. The correct option is D. Note that this differs from the answer key, but the solution explicitly states option D and computes $H' = 16H$.

Given: A resistor develops thermal energy H in $10 \, \text{s}$ when $I_1 = 4 \, \text{A}$.

Find: New thermal energy for $I_2 = 16 \, \text{A}$ in the same time.

Use Joule's law of heating:

$$E = I^2Rt$$

First case:

$$E_1 = I_1^2Rt$$

Second case:

$$E_2 = I_2^2Rt$$

Since $R$ and $t$ remain constant,

$$\frac{E_2}{E_1} = \frac{I_2^2}{I_1^2}$$

Substitute the values:

$$\frac{E_2}{E_1} = \left(\frac{16}{4}\right)^2 = 16$$

So,

$$E_2 = 16E_1 = 16H$$

Therefore, the thermal energy developed in $10 \, \text{s}$ is 16H.