Physics Q35 – JEE-MAIN 2025

Home Physics Electrostatics & CapacitanceQuestion #133

Three parallel plate capacitors

C_{1}

C_{2}

and

C_{3}

each of capacitance

5 μ F

are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of

C_{1}

capacitor is filled with a dielectric medium having dielectric constant of 4, is:

Clue 1 of 3: Circuit Topology

Identify the series and parallel combinations of capacitors in the given circuit diagram.

🥷

Ninja Strategy•Upper Bound Estimation

Estimate an upper bound for the total capacitance by recognizing that a series combination of capacitors always results in a capacitance smaller than the smallest individual capacitor in that series. This helps eliminate options that are too large.

🧠 Full Solution Path

Step 1: Determine the modified capacitance of

C_{1}

and identify the circuit configuration.✦ Active

Initially, all capacitors have a capacitance of $C = 5 μ F$ . When a dielectric medium with dielectric constant $K = 4$ is introduced into capacitor $C_{1}$ , its capacitance changes to $C_{1}^{'} = K C_{1}$ .

C_{1}^{'} = 4 \times 5 μ F = 20 μ F

From the figure, capacitors $C_{1}^{'}$ and $C_{2}$ are connected in series. This series combination is then connected in parallel with capacitor $C_{3}$ between points A and B.

Step 2: Calculate the equivalent capacitance of the series combination of

C_{1}^{'}

and

C_{2}

.○ Expand

The equivalent capacitance for $C_{1}^{'}$ and $C_{2}$ in series, denoted as $C_{12}$ , is given by:

\frac{1}{C_{12}} = \frac{1}{C_{1}^{'}} + \frac{1}{C_{2}} \frac{1}{C_{12}} = \frac{1}{20 μ F} + \frac{1}{5 μ F} \frac{1}{C_{12}} = \frac{1 + 4}{20 μ F} = \frac{5}{20 μ F} = \frac{1}{4 μ F} C_{12} = 4 μ F

Step 3: Calculate the total effective capacitance between points A and B.○ Expand

The equivalent capacitance $C_{12}$ is in parallel with $C_{3}$ . The total effective capacitance $C_{e q}$ between A and B is:

C_{e q} = C_{12} + C_{3} C_{e q} = 4 μ F + 5 μ F C_{e q} = 9 μ F

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Physics Question 35 – JEE-MAIN 2025

Clue 1 of 3: Circuit Topology

🧠 Full Solution Path