Question

A voltage Ee^-at is applied at t=0 to a circuit containing inductance L and resistance R. Determine the current at any time t.

(a) $\frac{Ee^{-at}}{L-Ra} \left(e^{\frac{-Rt}{L}}\right)$

(b) $\frac{E}{R-La} \left(e^{-at}-e^{\frac{-Rt}{L}}\right)$

(c) $\frac{E}{La} \left(e^{-at}+e^{\frac{Rt}{L}}\right)$

(d) $\frac{Ee^{-at}}{R} \left(1-e^{\frac{-Rt}{L}}\right)$

I had been asked this question during an online exam.

My question is based upon Simple Electrical Networks Solution in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct answer is (b) $\frac{E}{R-La} \left(e^{-at}-e^{\frac{-Rt}{L}}\right)$

The explanation: The differential equation for the circuit is derived based on Kirchhoff’s voltage law and it is given by $L \frac{di}{dt} + Ri = E \rightarrow \frac{di}{dt} + \frac{R}{L} \,i = \frac{Ee^{-at}}{L}$ this is of the form $\frac{dy}{dx} + Px = Q$

its solution is $ie^{\frac{Rt}{L}} = \int e^{\frac{Rt}{L}} * \frac{Ee^{-at}}{L} \,dt + c = \frac{E}{L}  \int e^{(\frac{R}{L}-a)t} \,dt + c$

$ie^{\frac{Rt}{L}} = \Big\{\frac{E}{L} * e^{(\frac{R}{L}-a)t} * \frac{1}{(\frac{R}{L}-a)}\Big\} + c = \frac{E}{R-La} * e^{(\frac{R}{L}-a)t} + c$

$i = \frac{Ee^{-at}}{L-Ra} + ce^{\frac{-Rt}{L}}$….(1) using i(0)=0 we get $c=\frac{-E}{R-La}$  

substituting in (1)

$i(t) =  \frac{E}{R-La} \left(e^{-at}-e^{\frac{-Rt}{L}}\right)$.