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CGP EDU Academic Team
Published on: September 12, 2026
A condenser having a capacity of 6 μ μ F is charged to 100 V and is then joined to an uncharged condenser of $14 \mu F$ and then removed. The ratio of the charges on 6 μ μ F and 14 μ μ F and the potential of 6 μ μ F will be
Text Solution
Verified by ExpertsThe correct answer is:
C
Let $q_1, q_2$ be the charges on two condensers
∴ ∴ $V = \frac{q_1}{6} = \frac{q_2}{14}$ ⇒ ⇒ $\frac{q_1}{q_2} = \frac{6}{14} = \frac{3}{7}$
Also $q_1 + q_2 = 600$ ⇒ ⇒ $q_1 + \frac{14}{6} q_1 = 600$
⇒ ⇒ $q_{1} = \frac{600}{20} \times 6$
∴ ∴ $v = \frac{q_1}{6} = \frac{600}{20} = 30 \text{ volt}$
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