Question

When the current in the portion of the circuit shown in the figure is $2\text{A}$ and increasing at the rate of $1\text{A/s}$, the measured potential difference $V_{ab}=8\text{V}$. However when the current is $2\text{A}$ and decreasing at the rate of $1\text{A/s}$, the measured potential difference $V_{ab}=4\text{V}$. The value of $R$ and $L$ are

• A. $3\Omega$ and $2\text{H}$ respectively
• B. $2\Omega$ and $3\text{H}$ respectively
• C. $10\Omega$ and $6\text{H}$ respectively
• D. $6\Omega$ and $1\text{H}$ respectively.

Answer 1

The correct option is A $3\Omega$ and $2\text{H}$ respectively

$V_{ab}=iR+L\frac{di}{dt}$
Now, $8=2R+L...\left(i\right)$
and $4=2R-L...\left(ii\right)$
Solving Eq. $\left(i\right)$ and Eq. $\left(ii\right)$, we get
$R=3\Omega$
and $L=2\text{H}$