Question

An inductor (L=0.03H) and a resistor (R=0.15 K$\Omega$) 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

$i_{\circ }=\frac{V}{R}=\frac{15v}{150\Omega }=.1A$

As the switch K1 is opened and K2 is closed The Inductor will behave as source and current at any time can be found using KCL

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

Integrating the equation (i)

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

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

In LR decay circuit the current is given as

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

at time t$=1$ ms current can be found by substituting $L=.03$ H

$R=150\Omega$

$i=.67mA$