Capacitors

JEE Main level, single correct, medium

10 of 180

Question 10 of 180

In the network shown below, the charge accumulated in the capacitor in steady state will be:

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Answer: (D) 7.2μC7.2\mu C

In steady state, capacitor behaves as open circuit, so no current will flow.
The total resistance of the circuit is, R=(6+4)Ω=10ΩR=\left(6+4\right)\Omega =10\Omega
The current flowing through the circuit is
I=3V10ΩI=\dfrac{3V}{10\Omega }
Potential difference on 6Ω6\Omega resistor
V′=310×6=1.8VV'=\dfrac{3}{10}\times 6=1.8V
The charge on the capacitor is,
Q=CV′=4μF×1.8V=7.2μCQ=CV'=4\mu F\times 1.8V=7.2\mu C

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