Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minQB

PhysicsMediumnumerical

In the following circuit $C_{1} = 12 μ F, C_{2} = C_{3} = 4 μ F$ and $C_{4} = C_{5} = 2 μ F$ . The energy stored in $C_{1}$ is $____$ $μ J$ .

Answer:

96.00

Solution:

Potential difference across the terminals of $C_{3}$ is 2 V.

$∴ Q_{3} = C V = (4 μ) (2) = 8 μ C$

Energy stored in $C_{1} = \frac{1}{2} \times 12 \times (4)^{2} = 96 μ J$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: QB

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