The network shown in figure is a part of a complex network. When current is 5A and is decreasing at the rate of 10³ A/s, then V_B – V_A will be

15V

Voltage across inductance

                  $ V = -L\frac{di}{dt} = 5\times 10^{-3}\times -10^3 = 5V$

            Since current is decreasing hence voltage across inductor will be positive.

Now Applying Kirchoff law between A and B , V_A – (5 x 1) + 15 + 5 = V_B

            Or  V_B – V_A = 15V