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A series RCL circuit is at resonance and contains a variable resistor that is set to 206Ω. The power dissipated in the circuit is 1.30 W. Assuming that the voltage remains constant, how much power is dissipated when the variable resistor is set to 532Ω? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

User GMS
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1 Answer

3 votes

Answer:

Power dissipated in resistor 532 ohm is 0.503 watt

Step-by-step explanation:

We have given in first case resistance
R_1=206ohm

Power dissipated in this resistance is
P_1=1.30watt

Power dissipated in the resistor is equal to
P=\frac{v_(rms)}^2{R}

We have to find the power dissipated in the resistor is 1.30 watt

From the relation we can say that
(P_1)/(P_2)=(R_2)/(R_1)


(1.3)/(P_2)=(532)/(206)


P_2=0.503watt

So power dissipated in resistor 532 ohm is 0.503 watt

User Srikar Kulkarni
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