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A series circuit has a 100-Ω resistor, 4.00-mH inductor and a 0.100-μF capacitor connected across a 120-V rms ac source at the resonance frequency. What is the power dissipated by the circuit

Respuesta :

Answer:

144 watt

Explanation:

resistance, R = 100 ohm

L = 4 mH

C = 100 micro farad

At resonance, the impedance is equal to R

Z = R

Vrms = 120 V

Irms = Vrms / R = 120 / 100 = 1.2 A

Power is given by

P = Vrms x Irms x CosФ

Where, CosФ is called power factor

At resonance, CosФ = 1  

Power, P = Irms x Vrms

P = 1.2 x 120 x 1

P = 144 Watt.

Thus, the power is 144 watt.

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