JEE MAIN - Physics (2015 (Offline) - No. 4)
An inductor $$(L=0.03$$ $$H)$$ and a resistor $$\left( {R = 0.15\,k\Omega } \right)$$ are connected in series to a battery of $$15V$$ $$EMF$$ 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 simultaneously. At $$t=1$$ $$ms,$$ the current in the circuit will be : $$\left( {{e^5} \cong 150} \right)$$
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$$6.7$$ $$mA$$
$$0.67$$ $$mA$$
$$100$$ $$mA$$
$$67$$ $$mA$$
Explanation
$${\rm I}\left( 0 \right) = {{15 \times 100} \over {0.15 \times {{10}^3}}} = 0.1A$$
$${\rm I}\left( \infty \right) = 0$$
$${\rm I}\left( t \right) = \left[ {{\rm I}\left( 0 \right) - {\rm I}\left( \infty \right)} \right]\,{e^{{{ - t} \over {L/R}}}} + i\left( \infty \right)$$
$${\rm I}\left( t \right) = 0.1\,{e^{{{ - t} \over {L/R}}}} = 0.1{e^{{R \over L}}}$$
$${\rm I}\left( t \right) = 0.1\,\,{e^{{{0.15 \times 1000} \over {0.03}}}} = 0.67mA$$
$${\rm I}\left( \infty \right) = 0$$
$${\rm I}\left( t \right) = \left[ {{\rm I}\left( 0 \right) - {\rm I}\left( \infty \right)} \right]\,{e^{{{ - t} \over {L/R}}}} + i\left( \infty \right)$$
$${\rm I}\left( t \right) = 0.1\,{e^{{{ - t} \over {L/R}}}} = 0.1{e^{{R \over L}}}$$
$${\rm I}\left( t \right) = 0.1\,\,{e^{{{0.15 \times 1000} \over {0.03}}}} = 0.67mA$$
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