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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisEasy2 minQB

PhysicsEasysingle choice

A $10 μF$ capacitor is fully charged to a potential difference of $50 V$ After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20 V$ . The capacitance of the second capacitor is :

Options:

Answer:

Solution:

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
$20 = \frac{10 \times 50 + 0}{20 + C}$
$C = 15 μ F$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: QB

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