In the circuit shown, charge on the 5 μ F capacitor is: (a) 5.45 μ C (b) 18.00 μ C (c) 10.90 μ C (d) 16.36 μ C

Answer: (d). Take the bottom middle node at 0 V; both cells put the left and right corners at +6 V. Let the top junction be at x. Total charge on its plates is zero: 2(x-6) + 4(x-6) + 5(x-0) = 0. 11x = 36 ⇒ x = 36/11 V Q₅ = 5×36/11 = 180/11 ≈ 16.36 So the answer is (d) 16.36 μ C.

In the circuit shown, charge on the 5 μ F capacitor is: | JEE Main 2020 (5 Sep, Shift 2)

Q73JEE Main 2020 · 5 SEP · SHIFT 2Medium

Combination of Capacitors

In the circuit shown, charge on the 5 $\mu$ F capacitor is:

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Answer: (d)

Take the bottom middle node at 0 V; both cells put the left and right corners at $+6$ V.
Let the top junction be at $x$ . Total charge on its plates is zero: $2(x-6) + 4(x-6) + 5(x-0) = 0$ .
$11x = 36 \Rightarrow x = \frac{36}{11}$ V
$Q_5 = 5\times\frac{36}{11} = \frac{180}{11} \approx 16.36\ \mu\text{C}$
So the answer is (d) 16.36 $\mu$ C.

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Combination of Capacitors question from JEE Main 2020 (5 Sep, Shift 2)

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