An inductor (L=0.03H) and a resistor (R=0.15k`Omega`) are connected in series to a battery of 15 V emf in a circuit shown below. The key `K_1` has bee

Question

An inductor (L=0.03H) and a resistor (R=0.15k`Omega`) are connected in series to a battery of 15 V emf in a circuit shown below. The key `K_1` has bee

An inductor (L=0.03H) and a resistor (R=0.15k`Omega`) are connected in series to a battery of 15 V emf in a circuit shown below. The key `K_1` has been kept closed for a long time. Then at t=0 , `K_1` is opened and key `K_2` is closed simulatenously. At t=1 ms, the current in the circuit will be `(e^5cong150)`
A. 100 mA
B. 67 mA
C. 6.7mA
D. 0.67 mA

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
When the key `K_1` is closed , current through the inductor,
`I_0=e/R=15/(0.15times10^3)=0.1A`
Here, t=1 ms =`10^-3` s
Time constant of the LR circuit,
`t_0=L/R=0.03/(0.15times10^3)=2times10^-4s`
`thereforet/t_0=10^-3/(2times10^-4)=5`
`thereforeI=I_0e^(-t//t_0)=0.1e^-5=0.1/150=0.67times10^-3A`
=0.67 mA