JEE PYQ: Capacitance Question 27
Question 27 - 2019 (09 Apr Shift 1)
Determine the charge on the capacitor in the following circuit:
(Circuit: 72 V battery, 6 $\Omega$ and 4 $\Omega$ in one branch, 2 $\Omega$ and 10 $\Omega$ in another, 10 $\mu$F capacitor across the 10 $\Omega$ resistor)
(1) 60 $\mu$C
(2) 2 $\mu$C
(3) 10 $\mu$C
(4) 200 $\mu$C
Show Answer
Answer: (4)
Solution
At steady state, there is no current in capacitor. $2\Omega$ and $10\Omega$ are in series (= $12\Omega$). This $12\Omega$ is parallel with $4\Omega$, combined $= \frac{12 \times 4}{12+4} = 3\Omega$. This is in series with $6\Omega$, total $= 9\Omega$. Current from battery $= \frac{72}{9} = 8$ A. Current in $10\Omega$: $i’ = \frac{4}{4+12} \times 8 = 2$ A. Pd across capacitor = $V = i’ \times R = 2 \times 10 = 20$ V. Charge $q = CV = 10 \times 20 = 200,\mu$C.
BETA
