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 r
esonance frequency. What is the power dissipated by the circuit
1 answer:
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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