A 3.5 mH inductor and a 4.5 mH inductor are connected in series and a time varying current is established in them. When the total emf of the combination is 16 V, the emf of the larger inductor is:
A) 7.0 V
B) 9.0 V
C) 2.3 V
D) 28 V
E) 36 V

Question

Quizplus · Accepted Answer

The total inductance of the circuit is 3.5 mH + 4.5 mH = 8 mH. Let the current in the circuit be I. From Faraday's law of electromagnetic induction, the emf induced in an inductor is given by:

emf = -L (dI/dt)

where L is the inductance of the inductor.

For the 3.5 mH inductor, the emf induced is -3.5×10^-3 (dI/dt). For the 4.5 mH inductor, the emf induced is -4.5×10^-3 (dI/dt).

Since the inductors are in series, the emfs add up to give the total emf of 16 V:

-3.5×10^-3 (dI/dt) - 4.5×10^-3 (dI/dt) = 16

Simplifying this equation, we get:

-8×10^-3 (dI/dt) = 16

or

dI/dt = -2×10^3 A/s

Now, let's find the emf induced in the 4.5 mH inductor:

emf = -4.5×10^-3 (dI/dt) = -4.5×10^-3 (-2×10^3) = 9 V

Therefore, the emf of the larger inductor is 9 V. Hence, the correct answer is B.

A 35 MH Inductor and a 4