Two identical inductors are connected in two different configurations P and Q, where a time varying current I(t) is flowing, as shown in the figure. The induced emf between points a and b for configuration P is E_P and that for configuration Q is E_Q. The ratio E_P/E_Q is : [Neglect the effect of mutual inductance.]
A. 2
Let the inductance of each identical inductor be L. For configuration (P): The two inductors are connected in series. The points a and b are located across the first inductor only. The current flowing through this inductor is I(t). The induced emf between points a and b is given by: E_P = L dI dt For configuration (Q): The two inductors are connected in parallel. The points a and b are the common nodes of this parallel combination. The equivalent inductance L_ eq between a and b is: L_ eq = L L L + L = L 2 The total current entering the parallel combination is I(t). The induced emf between points a and b is: E_Q = L_ eq dI dt = L 2 dI dt Now, finding the ratio of the induced emfs: E_P E_Q = L dI dt L 2 dI dt = 2 Answer: 2
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