First slide

Capacitance

Question

Capacitor A is charged to a potential of 100 V and, capacitor B is charged to a potential of 75 V. What are the charges on A and B after key is closed Fig.

Easy

A

$\frac{250}{3} μC; \frac{500}{3} μC$
B

$\frac{250}{4} μC; \frac{250}{3} μC$
C

$\frac{500}{4} μC; \frac{500}{3} μC$
D

$\frac{1250}{4} μC; \frac{1250}{3} μC$

Solution

$\begin{aligned} q_{A} = 5 \times 100 = 500 μC \\ q_{B} = 10 \times 75 = 750 μC \end{aligned}$
Net charge $= 750 μC - 500 μC = 250 μC$
( $∵$ Negative plate of A is connected to the positive plate of B)
Common potential, $V = \frac{250}{5 + 10} = \frac{250}{15} volt$
Now, $q_{A}^{'} = C_{1} V = 5 \times \frac{250}{15} = \frac{250}{3} μC$
and $q_{B}^{'} = C_{2} V = 10 \times \frac{250}{15} = \frac{500}{3} μC$

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