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 V_{B} – V_{A} 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 $