A part of complete circuit is shown in the figure. At some instant, the value of current I is 1 A and it is decreasing at a rate of 102As−1. The value of the potential difference VP−VQ, (in volts) at that instant, is ____.
Answer: 33
Solution
Trace the potential from Q to P through the branch. Crossing the resistor against the current drops IR=1×2=2V, the cell of emf 30V raises the potential by 30V, and the inductor contributes −LdtdI. Since the current is falling at dtdI=−102A/s with L=50×10−3H, this term is −(50×10−3)(−102)=+5V. Adding these, VP−VQ=−2+30+5=33V.