Q. The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5A , and is decreasing at a rate 10³ A/s then VB – VA is
(a) 20V
(b) 15V
(c) 10V
(d) 5V
Ans: (b)
Sol: $\displaystyle V_A – V_B = 5\times 1 – 15 -L\frac{di}{dt}$
$ \displaystyle V_A – V_B = 5\times 1 – 15 – 5 \times 10^{-3}\times 10^3$
$ \displaystyle V_A – V_B = -15 $
$ \displaystyle V_B – V_A = 15 $