The network shown in the figure is part of a complete circuit instant the current (I) is 5A and is decreasing at a rate of `10^(3)" A s"^(-1)`, then the value of `V_(B)-B_(A)` is

Options

A. 5 V
B. 10 V
C. 15 V
D. 20 V

Answer

C

Explanation

`V_(A)=iR+E-L(di)/dt=V_(B)`
`rArrV_(A)-V_(B)=5+15-5xx10^(-3)xx10^(3)=15V`.