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PhysicsEMI/ACSeries AC CircuitsMedium2 minPYQ_2020

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A series $L - R$ circuit is connected to a battery of emf $V$ . If the circuit is switched on at $t = 0$ , then the time at which the energy stored in the inductor reaches $(\frac{1}{n})$ times of its maximum value, is :

Options:

Answer:

Solution:

Potential energy stored in inductor is given by $U = \frac{1}{2} L^{2}$

$\cup \propto l^{2}$

$\frac{U}{U_{0}} = {(\frac{1}{I_{0}})}^{2}$

$\frac{1}{n} = {(\frac{1}{I_{0}})}^{2}$

$\frac{1}{I_{0}} = 1 - e^{- RT / L} = \frac{1}{\sqrt{n}}$

$t = \frac{L}{R} \ln \frac{\sqrt{n}}{\sqrt{n} - 1}$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2020

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