Question

a series circuit consists of a 50-Hz source, a 50-ohm resistor, a 0.50-H inductor, and a 60 microfarad capacitor. The rms current in the circuit is measured to be 1.3 Amps. What is the voltage amplitude of the source?

Answer 1

Answer:

The voltage amplitude of the source is 212.25 V

Explanation:

Given;

frequency of the circuit, F = 50-Hz

resistance across the circuit, R = 50-ohm

capacitance, C = 60 microfarad

inductance, L = 0.50-H

rms current in the circuit, $I_{rms}$ =  1.3 A

$I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_o = I_{rms} *\sqrt{2}$

where;

$I_o$ is peak current

$I_o = 1.3*\sqrt{2} = 1.8385 \ A$

Also, the voltage amplitude of the source is given as;

$V_o = I_oZ$

where;

Z is impedance calculated as;

$Z = \sqrt{(X_l-X_c)^2 + R^2}$

$X_l$ is inductive reactance, calculated as;

$X_l = \omega L = 2\pi f L = 2\pi * 50* 0.5 = 157.1 \ ohms$

$X_c$ is capacitive reactance, calculated as;

$X_c = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2 \pi *50*60*10^{-6}} = 53.0448 \ ohms$

$Z = \sqrt{(157.1 -53.0448)^2+50^2} = \sqrt{13327.4847} = 115.45\ ohms$

Finally, the voltage amplitude of the source:

$V_o = I_oZ\\\\V_o = 1.8385*115.45 = 212.25 \ V$

Therefore, the voltage amplitude of the source is 212.25 V