Question

A resistance of $100\Omega$ and inductance $L$ henry are connected in series with a battery of $20$ volts.The current at any instant, if the relation between is $L,R,E$ is $L\frac{di}{dt}+Ri=E$ is given by?

• A. $i=0.6(-e^{-200t})$
• B. $i=0.4(-e^{-200t})$
• C. $i=0.2(1-e^{\frac{-100}{L}t})$
• D. $i=0.8(-e^{-200t})$

Answer 1

The correct option is C $i=0.2(1-e^{\frac{-100}{L}t})$

$R=100,E=20$

$L\frac{di}{dt}+iR=E$

$\frac{di}{dt}+i\frac{R}{L}=\frac{E}{L}$

$IF=e^{\int \frac{R}{L}dt}=e^{\frac{R}{L}t}$

Therefore, we get

$d\left(e^{\frac{R}{L}t}i\right)=\int \frac{E}{L}{e}^{\frac{R}{L}t}dt$

$i{e}^{\frac{R}{L}t}$=$\frac{E}{L}\frac{e^{\frac{R}{L}t}}{\frac{R}{L}}+C$

$e^{\frac{R}{L}t}i$=$\frac{E}{R}{e}^{\frac{R}{L}t}+C$

$i=\frac{E}{R}+C{e}^{\frac{-R}{L}t}$

Now at $t=0^{+}$,$i=0$,$R=100$,$E=20$, As current won't change suddenly, we get

$0=0.2+C$

$\Rightarrow C=-0.2$

$i=0.2(1-e^{\frac{-100}{L}t})$