Published by:
CGP EDU Academic Team
Published on: September 12, 2026
In the adjoining figure, four capacitors are shown with their respective capacities and the P.D. applied. The charge and the P.D. across the $4 \mu F$ capacitor will be
Text Solution
Verified by ExpertsThe correct answer is:
D
Total capacitance $\frac{1}{c} = \frac{1}{20} + \frac{1}{8} + \frac{1}{12}$ ⇒ ⇒ $c = \frac{120}{31} \mu F$
Total charge $Q = CV = \frac{120}{31} \times 300 = 1161 \, \mu C$
Charge, through 4 μ μ F condenser $= \frac{1161}{2} = 580 \, \mu \mathrm{C}$
and potential difference across it $= \frac{580}{4} = 145 \mathrm{V}$
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