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

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