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Roman55 [17]
4 years ago
11

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 curren

t in the circuit is measured to be 1.3 Amps. What is the voltage amplitude of the source?
Physics
1 answer:
Ann [662]4 years ago
5 0

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

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Answer:

W = 5760 J

Explanation:

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Replace with the data, and solve:

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