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LuckyWell [14K]
3 years ago
9

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
Physics
1 answer:
scZoUnD [109]3 years ago
6 0

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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In order to calculate the thermal energy, first let's calculate the power, using the formula:

P=\frac{V^2}{R}

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\begin{gathered} E=P\cdot t \\ E=1.62\cdot7.5\cdot60 \\ E=729\text{ J} \end{gathered}

In scientific notation, we have an energy of 7.3 * 10^2 J, therefore the correct option is the fourth one.

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