When the current in the portion of the circuit shown in the figure is 2A and increasing at the rae of 1A/s, the measured potential difference `V_(ab)=8V`. However when the current is 2A and decreasing at the rate of 1A/s, the measured potential difference `V_(ab)=4V`. The values of R and L are

Options

A. `3Omega` and 2 H respectively
B. `3Omega` and 3 H respectively
C. `2Omega` and 1 H respectively
D. `3Omega` and 1 H respectively

Answer

A

Explanation

From the figure, `V_(a)-iR-L(di)/(dt)=V_(b)`
`therefore" "V_(a)-V_(b)=iR+L(di)/(dt)`
According to given conditions,
8 = 2R + L ….(i)
4 = 2R - L …..(ii)
Solving these two equation, we get
`R=3Omega and L = 2H`