Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsMedium2 minPYQ_2021

PhysicsMediumnumerical

An inductor of $10 mH$ is connected to a $20 V$ battery through a resistor of $10 kΩ$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after $1 μs$ is $\frac{x}{100} mA$ . Then $x$ is equal to ______ . (Take $e^{- 1} = 0.37$ )

Answer:

74.00

Solution:

$I_{\max} = \frac{V}{R} = \frac{20 V}{10 KΩ} = 2 mA$

For $L R$ -decay circuit,

$I = I_{\max} e^{- R t / L}$

$I = 2 mA e^{\frac{- 10 \times 10^{3} \times 1 \times 10^{- 6}}{10 \times 10^{- 3}}}$

$I = 2 mA e^{- 1}$

$I = 2 \times 0.37 mA$

$I = \frac{74}{100} mA$

$x = 74$

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

⏱ 2mℹ️ Source: PYQ_2021

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