JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 19)
An inductor of 10 mH is connected to a 20V battery through a resistor of 10 k$$\Omega$$ 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 $$\mu$$s is $${x \over {100}}$$ mA. Then x is equal to ___________. (Take e$$-$$1 = 0.37)
Answer
74
Explanation
$${I_{\max }} = {V \over R} = {{20V} \over {10K\Omega }} = 2$$ mA
For LR - decay circuit
$$I = {I_{\max }}{e^{ - Rt/L}}$$
$$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$$
I = 2mA e$$-$$1
I = 2 $$\times$$ 0.37 mA
$$I = {{74} \over {100}}$$ mA
x = 74
For LR - decay circuit
$$I = {I_{\max }}{e^{ - Rt/L}}$$
$$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$$
I = 2mA e$$-$$1
I = 2 $$\times$$ 0.37 mA
$$I = {{74} \over {100}}$$ mA
x = 74
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