The network shown in figure is a part of a complete circuit. If at a certain instant the current 'J' is 5 A and is decreasing at the rate of `10^3 As^(-1)`, then `V_A - V_B` is :

Options

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

Answer

C

Explanation

`(di)/(dt) = -10^(3)` A/sec

i(t) = 5A
Apply KVL
`V_(A) - 5 xx 1 - 15V - 5 xx 10^(-3) xx ((di)/(dt)) - V_(B) = 0`
`V_(A) - 5 xx 1 - 15 - 5 xx 10^(-3) xx (-10^(3)) - V_(B) =0`
`V_(A) -V_(B) = 15 V`