Two identical inductors are connected in two different configurations P (series) and Q (parallel), where a time-varying current I(t) flows, as shown. The induced emf between points a and b for configuration P is E_P and for Q is E_Q. The ratio E_P/E_Q is: [Neglect mutual inductance.]
- A.(1)/(4)
- B.(1)/(2)
- C.1
- D.2✓
Correct Answer
(D) 2
Solution & Explanation
Key principle: for an inductor, the emf across it is L(dI/dt). The trick is reading exactly which coils the points a and b span in each diagram. Configuration P (series stack): the point b is the junction BETWEEN the two coils, so the path a→b crosses only ONE coil. Hence E_P=L(dI/dt). Configuration Q (parallel): the two coils are connected in parallel between a and b. With no mutual inductance, the parallel combination is L_Q=(L·L)/(L+L)=L/2, and the same total current I(t) flows in from a to b, so E_Q=L_Q(dI/dt)=(L/2)(dI/dt). Taking the ratio: E_P/E_Q=[L(dI/dt)]/[(L/2)(dI/dt)]=L/(L/2)=2. This matches option D (ratio = 2). Trap: if you wrongly took E_P across both series coils (2L), you would get E_P/E_Q=2L/(L/2)=4. The point b sitting at the mid-junction is what makes E_P span a single coil.
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