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elena-s [515]
3 years ago
12

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V . The rat

e at which electrical energy is dissipated in the resistor is 296 W Part A
What is the impedance Z of the circuit?
Express your answer to three significant figures and include the appropriate units.
Part B
What is the amplitude of the voltage across the inductor?
Express your answer to three significant figures and include the appropriate units
Part C
What is the power factor?
Express your answer using three significant figures.
Physics
1 answer:
max2010maxim [7]3 years ago
6 0

Answer:

Explanation:

 Rate of dissipation of energy at resistor

= I² R = 296

current I² = 296/R = 296 / 300

I = 0.9933 A.

Impedence Z = V / I = 500 / .9933

= 503 ohm.

B ) Let it be V

voltage across resistance = 300 x .9933 = 298 v

Now 500² = 298² + V²

V = 401 v

C)  Power factor = R / Z

=300 / 503

= 0.596

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