Question

Figure shows a circuit consisting of an ideal cell,an inductor ' L' and a resistor 'R', connected in series. Let switch 'S' be closed at t=0.Suppose at t=0 the current in the inductor is $i_0$, then find out the equation of current as a function of time?

Answer 1

Let at an instant t, the current in the circuit is i which is increasing at the rate of time , i.e. di/dt.
According to KVL rule along the circuit , we have
$\Rightarrow L\frac{di}{dt}=\epsilon -iR$

$\Rightarrow {\int }_{i_0}^i\frac{di}{\epsilon -iR}={\int }_0^t\frac{dt}{L}$

$\Rightarrow ln\left(\frac{\epsilon -iR}{\epsilon -i_{0}R}\right)=-\frac{Rt}{L}$

$\Rightarrow \epsilon -iR=(\epsilon -i_{0}R)e^{-Rt/L}$

$\Rightarrow i=\frac{\epsilon -(\epsilon -i_{0}R)e^{-Rt/L}}{R}$