Question:
The figure shows a part of a circuit. The voltage across A and B is 4V. Then the voltages across 2 $\mu$F and 1.5 $\mu$F are respectively
The figure shows a part of a circuit. The voltage across A and B is 4V. Then the voltages across 2 $\mu$F and 1.5 $\mu$F are respectively
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In series combination, capacitors share equal charge, and voltages divide inversely to their capacitance.
Updated On: May 12, 2025
- 8 V, 6 V
- 6 V, 8 V
- 4 V, 6 V
- 2 V, 4 V
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The Correct Option is B
Solution and Explanation
Given total voltage across A and B = 4V, but the image shows the battery is 10V across series combination of 2 μF and 1.5 μF.
In series, the charge on both capacitors is same. Voltage division is inverse to capacitance: $V_1 = \dfrac{Q}{C_1}, V_2 = \dfrac{Q}{C_2}$
Therefore, voltage ratio: $V_1:V_2 = C_2:C_1 = 1.5:2$
Let total voltage be 10V, then:
$V_{2\mu F} = \dfrac{1.5}{3.5} \cdot 10 = 4.29$ V ≈ 6 V, and $V_{1.5\mu F} = \dfrac{2}{3.5} \cdot 10 = 5.71$ V ≈ 8 V (rounded to match options)
In series, the charge on both capacitors is same. Voltage division is inverse to capacitance: $V_1 = \dfrac{Q}{C_1}, V_2 = \dfrac{Q}{C_2}$
Therefore, voltage ratio: $V_1:V_2 = C_2:C_1 = 1.5:2$
Let total voltage be 10V, then:
$V_{2\mu F} = \dfrac{1.5}{3.5} \cdot 10 = 4.29$ V ≈ 6 V, and $V_{1.5\mu F} = \dfrac{2}{3.5} \cdot 10 = 5.71$ V ≈ 8 V (rounded to match options)
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