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Q. In the adjoining circuit diagram, $E=5 \, V$ , $r=1 \, \Omega $ , $R_{2}=4 \, \Omega $ , $R_{1}=R_{3}=1 \, \Omega $ and $C=3 \, mF$ . Then the numerical value of the charge on each plate of the capacitor is

Question

NTA AbhyasNTA Abhyas 2020Current Electricity

Solution:

Solution
Current through resistors $R_{1}$ and $R_{3}$ in steady state is zero.
Therefore current through $R_{2}$ is
$i_{2}=\frac{E}{R_{2} + r}=\frac{5}{4 + 1}=1A$
Therefore charge on capacitor is
$q=C\left(\frac{i_{2} R_{2}}{2}\right)=6 \, μC$