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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACL-R circuitEasy2 minPYQ_2020

PhysicsEasysingle choice

An inductor of inductance $L = 400$ mH and resistors of $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery emf $12 V$ as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at $t = 0$ . The potential drop across L as a function of time is

Options:

Answer:

Solution:

In the branch containing L and $R_{2}$

$i = \frac{E}{R_{2}} (1 - e^{- \frac{R_{2} t}{L}})$

$\frac{d i}{d t} = \frac{E}{R_{2}} e^{- \frac{R_{2} t}{L}} . \frac{R_{2}}{L} = \frac{E}{L} e^{- \frac{R_{2} t}{L}}$

$∴ V_{L} = L \frac{d i}{d t} = E e^{- \frac{R_{2} t}{L}}$

$= 12 e^{- \frac{2 t}{400 \times 10^{- 3}}} = 12 e^{- 5 t} V$

Subject:PhysicsTopic:EMI/ACSubtopic:L-R circuit

⏱ 2mℹ️ Source: PYQ_2020

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