Question
An inductor $$\left( {L = 0.03\,H} \right)$$ 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 :
An inductor $$\left( {L = 0.03\,H} \right)$$ 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 :
$$\left( {{e^5} \cong 150} \right)$$
A.
$$6.7\,mA$$
B.
$$0.67\,mA$$
C.
$$100\,mA$$
D.
$$67\,mA$$
Answer :
$$0.67\,mA$$
Solution :
$$\eqalign{ & I\left( 0 \right) = \frac{{15 \times 100}}{{0.15 \times {{10}^3}}} = 0.1\,A \cr & I\left( \infty \right) = 0 \cr & I\left( t \right) = \left[ {I\left( 0 \right) - I\left( \infty \right)} \right]{e^{\frac{{ - t}}{{\frac{L}{T}}}}} + i\left( \infty \right) \cr & I\left( t \right) = 0.1\,{e^{\frac{{ - t}}{{\frac{L}{R}}}}} = 0.1\,{e^{\frac{R}{L}}} \cr & I\left( t \right) = 0.1\,e\frac{{0.15 \times 1000}}{{0.03}} = 0.67\,mA \cr} $$
$$\eqalign{ & I\left( 0 \right) = \frac{{15 \times 100}}{{0.15 \times {{10}^3}}} = 0.1\,A \cr & I\left( \infty \right) = 0 \cr & I\left( t \right) = \left[ {I\left( 0 \right) - I\left( \infty \right)} \right]{e^{\frac{{ - t}}{{\frac{L}{T}}}}} + i\left( \infty \right) \cr & I\left( t \right) = 0.1\,{e^{\frac{{ - t}}{{\frac{L}{R}}}}} = 0.1\,{e^{\frac{R}{L}}} \cr & I\left( t \right) = 0.1\,e\frac{{0.15 \times 1000}}{{0.03}} = 0.67\,mA \cr} $$
