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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle 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

Options:

Answer:

Solution:

Energy stored in capacitor is equal to energy stored in inductance

$\frac{1}{2} C V^{2} = \frac{1}{2} L I^{2}$

$C = \frac{L I^{2}}{V^{2}}$

$= \frac{1 \times {(2)}^{2}}{{(400)}^{2}}$

$= 25 μF$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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