Question:
Two identical air core capacitors are connected in series to a voltage source of 15 V. If one of the capacitors is filled with a medium of dielectric constant 4, the new potential across this capacitor is
Two identical air core capacitors are connected in series to a voltage source of 15 V. If one of the capacitors is filled with a medium of dielectric constant 4, the new potential across this capacitor is
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In series capacitors, \(Q\) is same for all. \(V \propto \dfrac{1}{C}\); higher capacitance means lower voltage.
Updated On: May 2, 2026
- 5 V
- 8 V
- 10 V
- 12 V
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
When capacitors are connected in series, the charge on each capacitor is the same. The voltage across each capacitor is \(V = Q/C\).
Step 2: Detailed Explanation:
Let the capacitance of each capacitor initially be \(C\). After inserting dielectric of constant \(K=4\), capacitance becomes \(4C\). For series combination: \[ Q = C_{eq} \times V_{total} = \frac{C \cdot 4C}{C + 4C} \times 15 = \frac{4C}{5} \times 15 = 12C \] Voltage across the air capacitor (capacitance \(C\)): \[ V_1 = \frac{Q}{C} = \frac{12C}{C} = 12 \text{ V} \] Voltage across dielectric-filled capacitor (capacitance \(4C\)): \[ V_2 = \frac{Q}{4C} = \frac{12C}{4C} = 3 \text{ V} \]
Step 3: Final Answer:
The new potential across the dielectric-filled capacitor is 3 V, and across the air capacitor is 12 V.
When capacitors are connected in series, the charge on each capacitor is the same. The voltage across each capacitor is \(V = Q/C\).
Step 2: Detailed Explanation:
Let the capacitance of each capacitor initially be \(C\). After inserting dielectric of constant \(K=4\), capacitance becomes \(4C\). For series combination: \[ Q = C_{eq} \times V_{total} = \frac{C \cdot 4C}{C + 4C} \times 15 = \frac{4C}{5} \times 15 = 12C \] Voltage across the air capacitor (capacitance \(C\)): \[ V_1 = \frac{Q}{C} = \frac{12C}{C} = 12 \text{ V} \] Voltage across dielectric-filled capacitor (capacitance \(4C\)): \[ V_2 = \frac{Q}{4C} = \frac{12C}{4C} = 3 \text{ V} \]
Step 3: Final Answer:
The new potential across the dielectric-filled capacitor is 3 V, and across the air capacitor is 12 V.
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