An ac generator produces 10 V (rms) at 400 rad/s. It is connected to a series RL circuit (R = 17.3 $\Omega$, L = 0.025 H). The rms current is:
A) 0.50 A and leads the emf by 30$\degree$
B) 0.71 A and lags the emf by 30$\degree$
C) 1.40 A and lags the emf by 60$\degree$
D) 0.50 A and lags the emf by 30$\degree$
E) 0.58 A and leads the emf by 90$\degree$

Question

Quizplus · Accepted Answer

We can use the formula for the impedance of a series RL circuit: 
Z = √(R^2 + (ωL)^2)
where ω is the angular frequency (ω = 2πf) and L is the inductance.

Plugging in the values, we get:
Z = √(17.3^2 + (400*0.025*2π)^2) = 22.14 Ω
(Note: we multiplied 0.025 H by 2π to convert from angular frequency to regular frequency.)

The current can then be calculated using Ohm's Law:
I = V/Z = 10/22.14 = 0.451 A (or 0.50 A, rounded to two significant figures)

The circuit contains only a resistor and an inductor, so the current lags the voltage by 90 degrees (or π/2 radians). Thus, the final answer is D, 0.50 A and lags the emf by 30°.

An Ac Generator Produces 10 V (Rms) at 400 Rad/s Ω\OmegaΩ

An Ac Generator Produces 10 V (Rms) at 400 Rad/s $\Omega$