An inductor (L =0.03 H) 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)` is opened and Key `K_(2)` is closed simultaneously. At t =1 ms, the current in the circuit will be `(e^(5) = 150)` A. `100 mA`B. `67 mA`C. `6.7 mA`D. `0.67 mA`

An inductor (L =0.03 H) and a resistor `(R = 0.15k(Omega))` are connec

1.	An inductor (L =0.03 H) 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)` is opened and Key `K_(2)` is closed simultaneously. At t =1 ms, the current in the circuit will be `(e^(5) = 150)` A. `100 mA`B. `67 mA`C. `6.7 mA`D. `0.67 mA`
Answer» Correct Answer - D `I= I_(0e)^((1)/(tau)), tau = (L)/(R )` `= (15)/(150) e^((-1 xx 10^(-3))/(1 //5 xx10^(3))) = 0.67 mA`

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