QJEEM24Ph4c5c: JEE Main 2024 Physics - Class 12 JEEM24Ph4c5c Solution: The charge accumulated on the capacitor connected in the…

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QJEEM24Ph4c5c: JEE Main 2024 Physics - Class 12 JEEM24Ph4c5c Solution: The charge accumulated on the capacitor connected in the…

The charge accumulated on the capacitor connected in the following circuit is _______ $μ C$ (Given $C = 150 μ F)$

JEE Main 2024 (Online) 27th January Morning Shift Physics - Capacitor Question 21 English

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JEE Main 2024 (Online) 27th January Morning Shift Physics - Capacitor Question 21 English Explanation

$\begin{aligned} V_{A} + \frac{10}{3} (1) - 6 (1) = V_{B} \\ V_{A} - V_{B} = 6 - \frac{10}{3} = \frac{8}{3} volt \\ Q = C (V_{A} - V_{B}) \\ = 150 \times \frac{8}{3} = 400 μ C \end{aligned}$

QJEEM24Ph4c5c: JEE Main 2024 Physics - Class 12 JEEM24Ph4c5c Solution: The charge accumulated on the capacitor connected in the…

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