Question

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by L $\frac{di}{dt}$+ R i = E. If E is constant and initially no current passes through the circuit, prove that $i=\frac{E}{R}\left\{1-e^{-\left(R/L\right)t}\right\}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ L\frac{di}{dt}+Ri=E \\ \Rightarrow \frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}.....\left(1\right) \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int \frac{R}{L}dt} \\ =e^{\frac{R}{L}t} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{\frac{R}{L}t},\mathrm{we}\mathrm{get} \\ {e}^{\frac{R}{L}t}\left(\frac{di}{dt}+\frac{R}{L}i\right)=e^{\frac{R}{L}t}\times \frac{E}{L} \\ \Rightarrow e^{\frac{R}{L}t}\frac{di}{dt}+e^{\frac{R}{L}t}\frac{R}{L}i=e^{\frac{R}{L}t}\times \frac{E}{L} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ {e}^{\frac{R}{L}t}i=\frac{E}{L}\int e^{\frac{R}{L}t}dt+C \\ \Rightarrow e^{\frac{R}{L}t}i=\frac{E}{L}\times \frac{L}{R}{e}^{\frac{R}{L}t}+C \\ \Rightarrow e^{\frac{R}{L}t}i=\frac{E}{R}{e}^{\frac{R}{L}t}+C.....\left(2\right) \\ \mathrm{Now}, \\ i=0\mathrm{at}t=0 \\ \therefore e^0\times 0=\frac{E}{R}{e}^0+C \\ \Rightarrow C=-\frac{E}{R} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ {e}^{\frac{R}{L}t}i=\frac{E}{R}{e}^{\frac{R}{L}t}-\frac{E}{R} \\ \Rightarrow i=\frac{E}{R}-\frac{E}{R}{e}^{-\frac{R}{L}t} \\ \Rightarrow i=\frac{E}{R}\left(1-e^{-\frac{R}{L}t}\right) \end{gathered}$$