Question

An inductor ($L=0.03H$) and a resistor ($R=0.15k\Omega$) 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 $t=1ms$, the current in the circuit will be $(e^5\cong 150)$

• A. $100mA$
• B. $67mA$
• C. $6.7mA$
• D. $0.67mA$

Answer 1

The correct option is D $0.67mA$

As the inductor will behave as zero resistance after infinite time in LR circuit, the current will be

$i_{\circ }=\frac{V}{R}=\frac{15V}{150\Omega }=0.1A$

As the switch $K_1$ is opened and $K_2$ is closed, the inductor will behave as a source and the current at any time can be found using KCL.

$-L\frac{di}{dt}-IR=0$ ....(i)

Integrating the equation (i)

$lo{g}_e\frac{i}{i_{\circ }}=-\frac{Rt}{L}$

$\Rightarrow i=i_{\circ }{e}^{\frac{-Rt}{L}}$

In LR decay circuit, the current is given as

$i=i_{\circ }{e}^{\frac{-Rt}{L}}=i_{\circ }{e}^{-5}$

At time $t=1ms$, current can be found by substituting $L=0.03H$

$R=150\Omega$

$i=0.67mA$