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, [tex]I_{rms}[/tex] = 1.3 A
[tex]I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_o = I_{rms} *\sqrt{2}[/tex]
where;
[tex]I_o[/tex] is peak current
[tex]I_o = 1.3*\sqrt{2} = 1.8385 \ A[/tex]
Also, the voltage amplitude of the source is given as;
[tex]V_o = I_oZ[/tex]
where;
Z is impedance calculated as;
[tex]Z = \sqrt{(X_l-X_c)^2 + R^2}[/tex]
[tex]X_l[/tex] is inductive reactance, calculated as;
[tex]X_l = \omega L = 2\pi f L = 2\pi * 50* 0.5 = 157.1 \ ohms[/tex]
[tex]X_c[/tex] is capacitive reactance, calculated as;
[tex]X_c = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2 \pi *50*60*10^{-6}} = 53.0448 \ ohms[/tex]
[tex]Z = \sqrt{(157.1 -53.0448)^2+50^2} = \sqrt{13327.4847} = 115.45\ ohms[/tex]
Finally, the voltage amplitude of the source:
[tex]V_o = I_oZ\\\\V_o = 1.8385*115.45 = 212.25 \ V[/tex]
Therefore, the voltage amplitude of the source is 212.25 V
