Question

An inductor $(L=0.03\text{H})$ and a resistor $(R=0.15k\Omega )$ are connected in series to a battery of $15\text{V}$ 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 $t=1\text{ms}$ , the current in the circuit will be : $e^5\cong 150$

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

Answer 1

Answer: (A)

Case I: ${\text{K}}_1$ is closed for long tiime

for long time, inductor acts as a conducting wire.
$\Rightarrow$ current in the circuit $=\frac{\text{V}}{\text{R}}$ $=\frac{15}{150}$
${\text{i}}_0=\text{0.1 A}$
Case II: ${\text{K}}_1$ is open and ${\text{K}}_2$ is closed

Current in the circuit
$i=i_0{e}^{\frac{-t}{\tau }};\tau =\frac{L}{R}$
After $t=1ms=10^{-3}s$
$i=i_0{e}^{-\left(\frac{10^{-3}\times 150}{3\times 10^{-2}}\right)}$
$=0.1e^{\frac{-15}{3}}=0.1\frac{1}{e^5}=\frac{0.1}{150}A=0.67mA$