The network shown in the figure is a part of a complete circuit. If at a certain instant the current i is 5 A and is decreasing at the rate of 103 A/s then VB-VA is
5 V
10 V
15 V
20 V
By using Kirchhoff's voltage law,
VA-iR+E-Ldidt=VB⇒VB-VA=-5×1+15-5×10-3×-103volt Therefore vB-vA=15 volt.