Question

A 2300-2 resistor and a 1.2-uF capacitor are connected in series across a generator (60.0 Hz, 120 V rms). Determine the average power delivered to the circuit. 4.514 Additional Materials Section 23.3

Answer 1

Solution)

Here,

Capacitive Reactance, XC = 1/ω* C

= 1/ 2 * π * f * C

= 1/ 2 * 3.14 * 60.0 * 1.2 * 10^-6

= 8846.42 Ω

Now,

Impedance Z = √(R2 + XC^2)

= √(2300^2 + 8846.42^2)

= 9140.52 Ω

Now,

Power factor , cosφ = R / Z

= 2300/ 9140.52

= 0.2516

Average power, Pavg = (Vrms^2/ Z) * cos φ

= 120^2* 0.25160 / 9140.52

= 0.396 W

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