Question

State whether the following statement is True or False?
The differential equation of a circuit with inductance $L$ and resistance $R$ is given by $\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}{e}^{-at}$ $(i=0att=0)$.The current at time $t$ is given by $i=\frac{E}{R-aL}\left(e^{-at}\right)$

• A. True
• B. False

Answer 1

The correct option is B False

We have $\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}{e}^{-at}$

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q$. Its solution is

$i\cdot e^{\int \left(R/L\right)dt}=\int e^{\int \left(R/L\right)dt}\cdot \frac{E}{L}{e}^{-at}dt+c$

$\therefore i\cdot e^{Rt/L}=\frac{E}{L}\int e^{\left(Rt/L\right)}\cdot e^{-at}dt+c=\frac{E}{L}\int e^{\left[\right(R/L)-a]t}dt+c$

$i\cdot e^{Rt/L}=\frac{E}{L}\cdot \frac{e^{\left[\right(R/L)-a]t}}{\left(R/L\right)-a}+c=\frac{E}{R-La}\cdot e^{\left[\right(R/L)-a]t}+c$

By data, when $t=0,i=0$ $\therefore c=-\frac{E}{R-La}$

$\therefore i\cdot e^{Rt/L}=\frac{E}{R-La}\cdot e^{\left[\right(R/L)-a]t}-\frac{E}{R-La}$

$\therefore i=\frac{E}{R-La}\cdot e^{-at}-\frac{E}{R-La}\cdot e^{-Rt/L}$

$=\frac{E}{R-La}(e^{-at}-e^{-Rt/L})$.