JEE Physics 2025 Question with Solution

Given: A capacitor of $8 \, \mu F$ is connected in parallel with the $10 \, \Omega$ resistor. The resistors $10 \, \Omega$ and $15 \, \Omega$ are in series across a $5 \, \text{V}$ battery.

Find: Charge on the capacitor at steady state.

At steady state in a DC circuit, the capacitor behaves as an open circuit. Therefore, the voltage across the capacitor becomes equal to the voltage across the $10 \, \Omega$ resistor.

The current through the resistive branch is

$$I = \frac{5}{10+15} = \frac{5}{25} = 0.2 \, \text{A}$$

So, the potential difference across the $10 \, \Omega$ resistor is

$$V_{10\Omega} = I \times 10 = 0.2 \times 10 = 2 \, \text{V}$$

Hence the capacitor voltage is

$$V_C = 2 \, \text{V}$$

Now use

$$Q = CV$$

Substituting,

$$Q = 8 \, \mu F \times 2 \, \text{V} = 16 \, \mu C$$

Therefore, the charge on the capacitor is $16 \, \mu C$.

Given: Series resistors $10 \, \Omega$ and $15 \, \Omega$ across $5 \, \text{V}$, with an $8 \, \mu F$ capacitor in parallel with the $10 \, \Omega$ resistor.

Find: Final charge on the capacitor.

Since the capacitor is in parallel with the $10 \, \Omega$ resistor, it must have the same steady-state voltage as that resistor.

Using voltage division,

$$V_C = V_{10\Omega} = 5 \times \frac{10}{10+15}$$ $$V_C = 5 \times \frac{10}{25} = 2 \, \text{V}$$

Then,

$$Q = CV = 8 \, \mu F \times 2 \, \text{V} = 16 \, \mu C$$

Therefore, the required numerical value is 16.

The solution states the final answer correctly as $16 \, \mu C$, although part of its intermediate explanation generically assumes the capacitor voltage equals the battery voltage, which does not match this circuit. The correct circuit-based working gives the same final value.