An RL series circuit is connected to an ac generator with a maximum emf of 20 V. If the maximum potential difference across the resistor is 16 V, then the maximum potential difference across the inductor is:
A) 2 V
B) 4 V
C) 12 V
D) 25.6 V
E) 36 V

Question

Quizplus · Accepted Answer

In an RL circuit, the potential difference across the resistor (VR) and the potential difference across the inductor (VL) can be found using the following equations:

VR = IR * R 
VL = IXL

where IR is the current through the resistor, R is the resistance of the resistor, and IXL is the current through the inductor multiplied by the inductive reactance (XL).

Since the circuit is connected to an AC generator, the current through the circuit is an alternating current that varies sinusoidally with time. The maximum current (Imax) can be found using Ohm's law:

Imax = Vmax / Z

where Vmax is the maximum emf of the generator and Z is the impedance of the circuit. For an RL circuit, the impedance is given by:

Z = sqrt(R^2 + Xl^2)

where Xl is the inductive reactance of the inductor.

We are given that the maximum potential difference across the resistor is VR = 16 V. This means that the maximum voltage drop across the inductor is:

VL = sqrt(Vmax^2 - VR^2)

Plugging in the given values, we get:

VL = sqrt(20^2 - 16^2) = sqrt(144) = 12 V

Therefore, the correct answer is C: 12 V.

An RL Series Circuit Is Connected to an Ac Generator