JEE Physics 2024 Question with Solution

Given: Initial capacitor capacitance is $C_1 = 2 \, \mu \text{F}$ and it is charged to $V_1 = 100 \, \text{V}$. The second capacitor has capacitance $C_2 = 6 \, \mu \text{F}$ and is initially uncharged.

Find: The common potential after connecting the two capacitors in parallel.

For an isolated system of connected capacitors, charge is conserved.

Initial charge on the charged capacitor:

$$Q = C_1 V_1 = 2 \, \mu \text{F} \times 100 \, \text{V} = 200 \, \mu \text{C}$$

Since the second capacitor is uncharged initially, total charge remains $200 \, \mu \text{C}$.

Total capacitance after parallel connection:

$$C_{\text{total}} = C_1 + C_2 = 2 \, \mu \text{F} + 6 \, \mu \text{F} = 8 \, \mu \text{F}$$

Hence the common potential is:

$$V_{\text{common}} = \frac{Q}{C_{\text{total}}} = \frac{200 \, \mu \text{C}}{8 \, \mu \text{F}} = 25 \, \text{V}$$

Therefore, the common potential across the capacitors is $25 \, \text{V}$. The correct option is B.

The solution discusses a different capacitor-energy question and concludes $x = 2$, so it does not match this question. The answer here is obtained from charge conservation for the given data.

Given: One capacitor of $2 \, \mu \text{F}$ is first charged by a $100 \, \text{V}$ battery. It is then connected in parallel with an uncharged capacitor of $6 \, \mu \text{F}$.

Find: The final common voltage.

First compute the charge stored on the first capacitor before disconnection:

$$Q_1 = C_1 V_1$$ $$Q_1 = 2 \, \mu \text{F} \times 100 \, \text{V} = 200 \, \mu \text{C}$$

Initially, the second capacitor is uncharged, so:

$$Q_2 = 0$$

After connection in parallel, both capacitors attain the same voltage $V$. Total charge is conserved:

$$Q_1 + Q_2 = (C_1 + C_2)V$$

Substitute the values:

$$200 \, \mu \text{C} = (2 \, \mu \text{F} + 6 \, \mu \text{F})V$$ $$200 \, \mu \text{C} = 8 \, \mu \text{F} \cdot V$$

So,

$$V = \frac{200}{8} = 25 \, \text{V}$$

Therefore, the final common potential is $25 \, \text{V}$, so the correct option is B.