Question

The time constant of an LR circuit is 40 ms. The circuit is connected at $t=0$ and the steady-state current is found to be 2.0 A. Find the current at (a) $t=10ms$ (b) $t=20ms,$ (c) $t=100ms$ and (d) $t=1s.$

Answer 1

Here $\tau =40ms,i_0=2A$

(a) $t=10ms,$

Since $i=i_0(1-e^{-t/\tau })$

$=2(1-e^{-10/40})$

$=2(1-e^{-1/4})$

$=2(1-0.7788)=2\left(0.2211\right)$

$=0.4422A=0.44A$

(b) When $t=20ms$

$i=i_0(1-e^{-t/\tau })$

$=2(1-e^{-100/40})$

$=2(1-e^{-10/4})$

$=2(1-0.082)$

$=1.835\cong 1.8A.$

(c) When $t=1s$

$i=i_0(1-e^{-t/\tau })$

$=2(1-e^{-1/40}\times {10}^{-3})$

$=2(1-e^{-100/4})$

$=2(1-e^{-25})=2\times 1=2A.$