JEE Physics 2025 Question with Solution

Given: $C_1 = C_2 = C_3 = 5 \, \mu\text{F}$ and dielectric constant for $C_1$ is $k = 4$.

Find: Effective capacitance between A and B.

When a dielectric is inserted in $C_1$, its capacitance becomes

$$C_1' = kC_1 = 4 \times 5 \, \mu\text{F} = 20 \, \mu\text{F}$$

Now $C_1'$ and $C_2$ are in series, so

$$\frac{1}{C_{12}} = \frac{1}{C_1'} + \frac{1}{C_2} = \frac{1}{20} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20}$$

Therefore,

$$C_{12} = 4 \, \mu\text{F}$$

This equivalent capacitance is in parallel with $C_3$, hence

$$C_{\text{total}} = C_{12} + C_3 = 4 \, \mu\text{F} + 5 \, \mu\text{F} = 9 \, \mu\text{F}$$

Therefore, the effective capacitance between A and B is $9 \, \mu\text{F}$. The correct option is C.

Given: Three capacitors each have capacitance $5 \, \mu\text{F}$. Only $C_1$ has dielectric constant $4$ inserted.

Find: Net capacitance across A and B.

First modify the capacitance of $C_1$:

$$C_1' = 4C_1 = 4 \times 5 \, \mu\text{F} = 20 \, \mu\text{F}$$

So the circuit now has:

• $C_1' = 20 \, \mu\text{F}$
• $C_2 = 5 \, \mu\text{F}$
• $C_3 = 5 \, \mu\text{F}$

From the figure, $C_1'$ and $C_2$ are in series on the top branch:

$$\frac{1}{C_{eq}} = \frac{1}{20} + \frac{1}{5}$$ $$\frac{1}{C_{eq}} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20}$$ $$C_{eq} = \frac{20}{5} = 4 \, \mu\text{F}$$

This top branch equivalent is in parallel with $C_3$:

$$C_{\text{total}} = C_{eq} + C_3 = 4 \, \mu\text{F} + 5 \, \mu\text{F} = 9 \, \mu\text{F}$$

Thus, the effective capacitance is $9 \, \mu\text{F}$, so the correct option is C.