The network shown in figure is a part of the circuit. (The battery has negligible resistance)

At a certain instant the current I = 2A and it is decreasing at the rate of `10^2 "As"^(-1)`. What is the potential difference between the points B and A ?

Options

A. 8.0 V
B. 8.5 V
C. 10 V
D. 15 V

Answer

B

Explanation

`V_A-IR+epsilon-(-L(dI)/(dt))=V_B`
`therefore V_B-V_A=-IR+ epsilon+L(dI)/(dt)`
`=-2xx2+12+5xx10^(-3)xx10^(-2)`
=-4+12+0.5
=8.5 V