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

At a certain instant the current I = 5A and is decreasing at a rate of `10^3 As^(-1)`. What is the potential difference between point B and A ?

Options

A. 5V
B. 10V
C. 15V
D. 0V

Answer

C

Explanation

Taking direction from A to B , the kirchoff.s law ,
`V_A-IR+epsilon-(-L(dI)/(dt))=V_B`
`therefore -IR+epsilon+(LdI)/(dt)=V_B-V_A`
Here `R=1 Omega , I=5A , (dI)/(dt) =10^3` A/s , L=5 mH, `epsilon`=15 V
`therefore -5xx1+15+5xx10^(-3)xx10^3=V_B-V_A`
`therefore -5+15+5=V_B-V_A`
`therefore V_B-V_A` =15 V