Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5A and it is decreasing at a rate of 10³ As^-1 if I is reversed in direction, then V_B - V_A equals 

In the previous question,

(A) 5 V

(B) 10 V

(C) 15 V

(D) 20 V

Answer 1

Correct option is (C) 15 V

When ‘i’ is reversed,

[i is decreasing against the direction of KVL. Hence \(\frac{di}{dt}\) = 10³].

V_A – V_B = – 5 – 15 + 5 

V_A – V_B = –15 V