A series circuit has a 50-Hz ac source, a 40-Ω resistor, a 0.70-H inductor, and a 60-μF capacitor, as shown in the figure. The rms current in the circuit is 1.6 A. What is the voltage amplitude of the source?

Question

Quizplus · Accepted Answer

Using Ohm's law, we can find the impedance of the circuit:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

where $X_L$ is the inductive reactance and $X_C$ is the capacitive reactance.

$X_L = 2\pi f L = 2\pi (50\,	ext{Hz})(0.70\,	ext{H}) = 220\,\Omega$

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (50\,	ext{Hz})(60\,	ext{\mu F})} \approx 53\,\Omega$

$Z = \sqrt{(40\,\Omega)^2 + (220\,\Omega - 53\,\Omega)^2} \approx 229 \,\Omega$

The voltage amplitude of the source can be found using Ohm's law:

$V_0 = IRMS\,Z = (1.6\,	ext{A})(229\,\Omega) \approx 366\,	ext{V}$

However, the question asks for the RMS voltage amplitude. Since the circuit is purely resistive, the RMS voltage amplitude is equal to the DC voltage that would produce the same power:

$P = \frac{V_{RMS}^2}{R} = \frac{(V_0/\sqrt{2})^2}{40\,\Omega} = \frac{V_0^2}{80\,\Omega}$

$P = I_{RMS}^2 R = (1.6\,	ext{A})^2 (40\,\Omega) = 102.4\,	ext{W}$

Setting these two expressions for power equal to each other, we can solve for $V_{RMS}$:

$\frac{V_0^2}{80\,\Omega} = 102.4\,	ext{W}$

$V_0 = \sqrt{102.4\,	ext{W} \cdot 80\,\Omega} \approx 390\,	ext{V}$

Therefore, the voltage amplitude of the source is 390 V.

A Series Circuit Has a 50-Hz Ac Source, a 40-Ω