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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisEasy2 minQB

PhysicsEasysingle choice

A $10 μF$ capacitor is charged to a potential difference of $50 V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 V$ . The capacitance of second capacitor is

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Solution:

$q_{1} = 10 \times 50 = 500 μC, C_{1} = 10 μF, C_{2} = ?, q_{2} = 0$

As, $V = \frac{q_{1} + q_{2}}{C_{1} + C_{2}}$

$C_{1} + C_{2} = \frac{q_{1} + q_{2}}{V} = \frac{500 + 0}{20} = 25 μF$

$C_{2} = 25 - C_{1} = 25 - 10 = 15 μF$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Force & Energy analysis

⏱ 2mℹ️ Source: QB

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