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
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a
5 V
b
10 V
c
15 V
d
20 V
answer is C.
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Detailed Solution
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.