Question

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current and the voltages in this circuit are not in phase with each other. Using the values given, the phase angle ϕ was found to be 73 ∘, and the current amplitude I was found to be 5.1×10−2 A . Calculate the power factor and the average power to the entire circuit and to each circuit element.

Answer 1

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Step-by-step explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

`Z=√(R^2+(X_L-X_C)^2)\\\Rightarrow Z=\sqrt{R^2+(\omega L-(1)/(\omega C))^2}\\\Rightarrow Z=\sqrt{280^2+(10500* 100* 10^(-3)-(1)/(10500* 0.8* 10^(-6)))^2}\\\Rightarrow Z=972.1483\ \Omega`

Power factor is given by

`F=(R)/(Z)\\\Rightarrow F=(280)/(972.1483)\\\Rightarrow F=0.28802`

The power factor is 0.28802

The average power to the circuit is given by

`P=(V^2)/(Z)\\\Rightarrow P=(50^2)/(972.1483)\\\Rightarrow P=2.57162\ W`

The average power to the circuit is 2.57162 W

Power to resistor

`P_R=IR\\\Rightarrow P_R=5.1* 10^(-2)* 280\\\Rightarrow P_R=14.28\ W`

Power to resistor is 14.28 W

Power to inductor

`P_L=IX_L\\\Rightarrow P_L=5.1* 10^(-2)* 10500* 100* 10^(-3)\\\Rightarrow P_L=53.55\ W`

Power to the inductor is 53.55 W

Power to the capacitor

`P_C=IX_C\\\Rightarrow P_C=5.1* 10^(-2)* (1)/(10500* 0.8* 10^(-6))\\\Rightarrow P_C=6.07142\ W`

The power to the capacitor is 6.07142 W