Question

The network shown in the figure is a part of a complete circuit. If at a certain instant, the current $i$ is $4\text{A}$ and is increasing at a rate of ${10}^3{\text{As}}^{-1}$. Then $V_B-V_A$ will be-

• A. $-11\text{V}$
• B. $-21\text{V}$
• C. $-31\text{V}$
• D. $-41\text{V}$

Answer 1

The correct option is B $-21\text{V}$

Given, $i=4\text{A};\frac{di}{dt}={10}^3{\text{As}}^{-1}R=1\Omega ;L=5\text{mH}$

Applying KVL to the given loop,

$V_A-(1\times i)-12-L\frac{di}{dt}$= $V_B$

$V_A-4-12-5\times {10}^{-3}\left({10}^3\right)$ = $V_B$

$V_B-V_A$= $-21\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.