Question

An inductor of inductance $2.0H$ and a resistor of resistance $10\Omega$ are connected in series to a battery of $EMF20V$ in a circuit as shown. The key $K_1$ has been kept closed for a long time. Then at $I=0,K_1$ is opened and key $K_2$ is closed simultaneously. The rate of decrease of current in the circuit at $t=1.0s$ will be $e^5=150$

• A. $\frac{1}{15}A/s$
• B. $\frac{2}{15}A/s$
• C. $\frac{1}{5}A/s$
• D. $\frac{4}{15}A/s$

Answer 1

The correct option is A $\frac{1}{15}A/s$

For $DC$ inductor offer $0$ resistance.

Hence, for $t<0,i=\frac{20}{1}=2Ampere$

Let current be $i\left(t\right)$

$V_R\left(t\right)=i\left(t\right)R$

$V_L\left(t\right)=\frac{Ldi\left(t\right)}{dt}$

$=\frac{Ldi\left(t\right)}{dt}+V_R\left(t\right)=0$

$\frac{Ldi\left(t\right)}{dt}+i\left(t\right)R=0$

$\frac{di\left(t\right)}{i\left(t\right)}=\frac{-R}{L}dt$

${\left[\ln i\left(t\right)\right]}_2^i={[-\frac{R}{L}t]}_0^t\ln \frac{i}{2}=-\frac{R}{L}ti=2e\frac{di}{dt}=2\times -\frac{R}{L}{e}^{-\frac{Rt}{L}},t=15=\frac{2R}{L}\times e^{-\frac{R}{L}}=2\times 5\times e^{-5}=\frac{10}{150}=\frac{1}{15}A/s$