Question

A constant electromotive force E volts is applied to a circuit containing a constant resistance R ohm in series with a constant inductance L henries. If the initial current is zero. What is the current in the circuit at any time t?

(a) \(\frac{E}{R} \left(1-e^{\frac{-Rt}{L}}\right)\)

(b) \(\frac{E}{R} \left(e^{\frac{Rt}{L}}\right)\)

(c) \(\frac{E}{L} (1-e^{RLt})\)

(d) \(\frac{E}{L} (e^{-RLt})\)

This question was addressed to me in examination.

I need to ask this question from Simple Electrical Networks Solution topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) \(\frac{E}{R} \left(1-e^{\frac{-Rt}{L}}\right)\)

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

i.e it is linear DE \( I.F = e^{\int P \,dx} = e^{\int \frac{R}{L} \,dt} = e^{\frac{Rt}{L}}, Q=\frac{E}{L} \) and its solution is given by

\(ye^{\int P \,dx} = \int Q e^{\int P \,dx} \,dx + c \rightarrow \,ie^{\frac{Rt}{L}} = \int e^{\frac{Rt}{L}} *  \frac{E}{L} \,dt + c\)

\(ie^{\frac{Rt}{L}} = (e^{\frac{Rt}{L}} * \frac{E}{L} * \frac{1}{R/L}) + c \rightarrow i = \frac{E}{R} + ce^{\frac{-Rt}{L}}\)….(1) given i(0)=0

\(\rightarrow -\frac{E}{R} = c\) substituting in (1) we get \(i(t) = \frac{E}{R}-\frac{E}{R} e^{\frac{-Rt}{L}} = \frac{E}{R} \left(1-e^{\frac{-Rt}{L}}\right).\)