Question

70. II The current through inductance $L$is given by$I=I_0{e}^{-t/\tau }$ .

CALC a. Find an expression for the potential difference $\Delta V_{\text{L}}$across the inductor.

b. Evaluate $\Delta V_{\text{L}}$at $t=0,1.0$, and $3.0\text{ms}$if $L=20\text{mH},I_0=50\text{mA}$, and $\tau =1.0\text{ms}$.

Short answer

Part (a) The potential difference is$\Delta V_L=\frac{L{I}_0}{\tau }{e}^{-\frac{t}{t}}$

Part (b) Potential difference at different times

time (S)	potential difference
0 s	1000 mV
1 s	370 mV
2 s	140 mV
3 s	50 mV

Step-by-step solution

Part (a)

The expression for current is,

$I=I_0{e}^{\frac{t}{r}}$

The expression for the potential difference,$\Delta V_L$is

$\begin{array}{l}\Delta V_L=-L\frac{dI}{dt}\\ \Delta V_L=-L\frac{d}{dt}\left(I_0{e}^{\frac{t}{t}}\right)\\ \Delta V_L=\frac{L{I}_0}{\tau }{e}^{-\frac{t}{t}}\dots \dots \left(1\right)\end{array}$

The potential difference is$\frac{L{I}_0}{\tau }{e}^{-\frac{t}{t}}$

Part (b) 

$L=20\text{mH}$,$\tau =1.0\text{ms}$ and$I_0=50\text{mA}$

At,$t=0s$

$\begin{array}{l}\Delta V_{L1}=\frac{\left(20\text{mH}\right)(50\text{mA})}{(1.0\text{ms})}{e}^{-0}\\ =1000\text{mV}\end{array}$

At, $t=1s$

$\Delta V_{L2}=\frac{\left(20\text{mH}\right)(50\text{mA})}{(1.0\text{ms})}{e}^{-1}=370\text{mV}$

At,$t=2s$

$\Delta V_{L3}(t=2s)=\frac{\left(20\text{mH}\right)(50\text{mA})}{(1.0\text{ms})}{e}^{-2}=140\text{mV}$

At,$t=3s$

$\Delta V_L=\frac{\left(20\text{mH}\right)(50\text{mA})}{(1.0\text{ms})}{e}^{-3}=50\text{mV}$