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Physics-

General

Easy

Question

A capacitor of capacitance $5 mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5 capital omega$ . The amount of charge on the capacitor plate is

$0 μ C$
$5 μ C$
$10 μ C$
$25 μ C$

The correct answer is: $10 mu C$

In steady state condition. No current flows through line (1). Hence total current $i equals fraction numerator 2.5 over denominator left parenthesis 1 plus 1 plus 0.5 right parenthesis end fraction equals 1 A$

Potential difference a cross line (2) = potential difference a cross capacitor
$equals 1 cross times 2 equals 2 V o l t$
So, charge on capacitor = 5 ´ 2 = 10 mC

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