Question

An electric circuit consisting of an electromotive force V, a resistance R and an inductor L. The current I at time t is given by:
I=(V/R(1-e^-Rt/L))

When the voltage is first applied (at t=0) the inductor opposes the rate of increase of current and I is small, however as t increases, I approaches V/R
(a) If L is the only independent variable, determine lim(I)
L 0
(b) If R is the only independent variable, determine lim(I)
R 0

Answer 1

Solution :-

$I= \frac{V}{R}(1-e^{-\frac{Rt}{L}})$

(a)

$L$ is the only independent variable here.

$\lim_{L \to 0}I=\lim_{L \to 0}\frac{V}{R}(1-e^{-\frac{Rt}{L}})$

As, $L \to 0 \Rightarrow \frac{1}{L}\to \infin \Rightarrow e^{-\frac{Rt}{L}} \to 0$

$\therefore \lim_{L \to 0}I=\lim_{L \to 0}\frac{V}{R}(1-e^{-\frac{Rt}{L}})$

$= \lim_{L \to 0}\frac{V}{R}(1-0)=\frac{V}{R}$

So, $\lim_{L \to 0}I =\frac{V}{R}$

(b)

$R$ is the only independent variable here.

$\lim_{R \to 0}I=\lim_{R \to 0}\frac{V}{R}(1-e^{-\frac{Rt}{L}})$

$= \lim_{R \to 0} \frac{V(0-(-t/L)e^{-\frac{Rt}{L}})}{1}$ [ Using L'Hospital ]

$= \lim_{R \to 0} \frac{Vte^{-\frac{Rt}{L}}}{L}=\frac{Vt}{L}$

So, $\lim_{R \to 0}I=\frac{Vt}{L}$