An ac generator producing 10 V (rms) at 200 rad/s is connected in series with a 50-$\Omega$ resistor, a 400-mH inductor, and a 200-$\mu$F capacitor. The rms voltage (in volts) across the resistor is:
A) 2.5
B) 3.4
C) 6.7
D) 10.0
E) 10.8

Question

Quizplus · Accepted Answer

We can use the concept of impedance to solve this problem.

The impedance of the resistor is simply its resistance:

$Z_R = R = 50 \Omega$

The impedance of the inductor is:

$Z_L = j\omega L = j(200)(0.4) = j80 \Omega$

The impedance of the capacitor is:

$Z_C = \frac{1}{j\omega C} = \frac{1}{j(200)(0.0002)} = -j25 \Omega$

The total impedance of the circuit is the sum of the individual impedances:

$Z_{total} = Z_R + Z_L + Z_C = 50 + j80 -j25 = 50 + j55 \Omega$

The current through the circuit is:

$I = \frac{V_{rms}}{Z_{total}} = \frac{10}{50 + j55} = 0.146 \angle -48.7^\circ$

The voltage across the resistor is simply:

$V_R = IZ_R = (0.146 \angle -48.7^\circ)(50) = 7.3 \angle -48.7^\circ$

Finally, the rms value of the voltage across the resistor is:

$V_{R,rms} = \frac{V_R}{\sqrt{2}} = \frac{7.3}{\sqrt{2}} = 5.2$

Therefore, the closest choice is C) 6.7.

An Ac Generator Producing 10 V (Rms) at 200 Rad/s Ω\OmegaΩ

An Ac Generator Producing 10 V (Rms) at 200 Rad/s $\Omega$