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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisMedium2 minPYQ_2014

PhysicsMediumsingle choice

A circuit has a self-inductance of $1 H$ and carries a current of $2 A$ . To prevent sparking, when the circuit is switched off, a capacitor which can withstand $400 V$ is used. The least capacitance of capacitor connected across the switch must be equal to

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Answer:

Solution:

Energy stored in capacitor $=$ energystored in inductance i.e., $\frac{1}{2} CV^{2} = \frac{1}{2} LI^{2}$ $\Rightarrow C = \frac{LI ^{2}}{V ^{2}} = \frac{1 \times ( 2 ) ^{2}}{( 400 ) ^{2}} = 25 μ F$

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

⏱ 2mℹ️ Source: PYQ_2014

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