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 EP and that for configuration Q is EQ. The ratio EP/EQ is: [Neglect the effect of mutual inductance.] [figure]
- 2
- 1/4
- ½
- 1
Answer
(1) 2
Solution
Let each inductor have self-inductance L.
In configuration P, current I(t) flows through the series combination, and points a and b are across one inductor:
EP = L | dIdt |
In configuration Q, the two identical inductors are in parallel between terminals a and b, so the equivalent inductance is Leq = L2:
EQ = Leq | dIdt | = L2 | dIdt |
Therefore, the ratio is:
EPEQ = LL/2 = 2
Answer: (1)