Question

The time constant of an LR circuit is 20 ms The circuit is connected at t = 0 and the steady-state current is found to be 4A. Find the current at 80 ms.

• A. 0.98 A
• B. 1 A
• C. 0.44 A
• D. 0.88 A

Answer 1

The correct option is D

0.88 A

Explanation for the correct answer:-

Step 1: Given Data

Time constant, $\tau =20ms$

$i_0=4A$

$t=80ms$

Step 2: Formula used,

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

$i$ is the circuit current.

$i_0$ is the steady state value of the current.

$\tau$ is the time constant.

t is the time taken

Step 3: Calculating the current at 80 ms

$i=i_0(1-e^{t/\tau })$ Given, $i_0$=4A at t=0, $t=80ms=80\times {10}^{-3}s$, $\tau =20ms=20\times {10}^{-3}s$

$i=4(1-e^{-\frac{\tau }{t}})$

$i=4(1-e^{-20\times {10}^{-3}/80\times {10}^{-3}})$

$$\begin{gathered} i=4(1-e^{-(1/4)}) \\ i=4(1-e^{-(0.25)}) \\ i=4(1-0.778) \\ i=0.88A \end{gathered}$$

Hence, option D is the correct answer.