Solve RLC Series Circuit: Find Vrms, Power Factor & Draw Phasor

In summary, the problem involved an RLC series circuit with a source of 141.4 Volts peak and a frequency of 1 kHz. The task was to find the RMS voltage over the resistor, the power factor, and draw a phasor diagram of the voltages. The RMS voltage over the inductor was 100 volts and the RMS voltage over the capacitor was 40 volts. After some confusion, it was determined that the voltage for the resistor should have been on the x-axis of the phasor diagram, rather than the voltage for the source.
  • #1
discoverer02
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1
I just finished an exam and this problem gave me fits. It seemed simple enough when I first looked at it, but it took me a while to come to the following conclusions which I'm not sure are correct.

The problem had an RLC series circuit with source of 141.4 Volts peak and a frequence of 1 kHz. The RMS voltage over the inductor is 100 volts and the RMS voltage over the capacitor is 40 volts.

I needed to find the RMS voltage over the resistor, the power factor and draw a phasor diagram of the voltages.

This is my reasoning and boy do I hope it's correct:

Since it's a series circuit the current is the same through all the components. Therefore, can't I use a phasor diagram to find Vrms over the resistor.

The voltage of the inductor would have a phase angle of 90 degrees and the voltage of the capacitor would have a phase angle of -90 degrees, so on the y-axis (complex axis) I'd have a vector 60 volts long and on the x-axis (real axis) I'd have the 100 rms volts from the source or would I have the unknown Vrms of the resistor?

Argh! I put the 100 volts rms for the source on the x-axis, but now I'm thinking it should have been the unknown Vrms over the resistor.

Too tired to think about out it right now, but I'd really appreciate some input.

Thanks
 
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  • #2
Yup, after considerable thought. The voltage for the resistor should have been on the x-axis, not the voltage for the source. I guess I was confused by the fact that you usually use the source voltage with a phase angle of 0 as your reference point.

Anyway it all comes back to me now. (too little too late for my exam).

Thanks anyway.
 
  • #3


Hi there,

First off, congratulations on finishing your exam! It's always tough to work through difficult problems, but it sounds like you put a lot of thought into this one.

From your description, it seems like you were on the right track with your reasoning. In a series circuit, the current is the same through all components, so you can use a phasor diagram to find the RMS voltage over the resistor. The voltage of the inductor will have a phase angle of 90 degrees (since it leads the current) and the voltage of the capacitor will have a phase angle of -90 degrees (since it lags the current). This means that the phasor for the inductor will be pointing up on the imaginary axis, and the phasor for the capacitor will be pointing down on the imaginary axis.

To find the RMS voltage over the resistor, you can use the Pythagorean theorem to find the length of the hypotenuse of the triangle formed by the phasor for the inductor and the phasor for the capacitor. This will give you the total RMS voltage in the circuit, which is equal to the RMS voltage over the resistor.

In this case, the total RMS voltage is 100 volts, so the RMS voltage over the resistor is also 100 volts. This means that the power factor is 1, since the voltage and current are in phase in a purely resistive circuit.

As for drawing the phasor diagram, you can put the 100 volts RMS for the resistor on the x-axis, since it represents the total RMS voltage in the circuit. The phasor for the inductor and the phasor for the capacitor will then be drawn relative to this 100 volt phasor, with the inductor's phasor pointing up and the capacitor's phasor pointing down on the imaginary axis.

I hope this helps and that you got the problem correct on your exam. Keep up the good work!
 

1. How do I find the value of Vrms in an RLC series circuit?

Vrms, or the root mean square voltage, can be calculated using the formula Vrms = Vmax/√2, where Vmax is the maximum voltage in the circuit. Alternatively, if you have the values for resistance (R), inductance (L), and capacitance (C), you can use the formula Vrms = IR√(R^2 + (ωL - 1/ωC)^2), where I is the current and ω is the angular frequency.

2. What is the power factor in an RLC series circuit?

The power factor in an RLC series circuit is a measure of how efficiently electrical power is being used. It is calculated by dividing the real power (P) by the apparent power (S), where P = Vrms x Irms and S = Vrms x Irms (taking into account the phase difference between voltage and current). A power factor of 1 indicates perfect efficiency, while a power factor less than 1 indicates some power loss.

3. How do I draw the phasor diagram for an RLC series circuit?

To draw the phasor diagram for an RLC series circuit, first identify the values of resistance (R), inductance (L), and capacitance (C). Then, draw a horizontal line to represent the voltage, and at one end of the line, draw a vertical line to represent the current. Use the values of R, L, and C to calculate the impedance (Z) of the circuit, and draw a diagonal line from the voltage line to the current line at an angle equal to the phase difference between voltage and current. Finally, use the values of R, L, and C to calculate the values of Vrms and Irms, and draw the respective lines at those values on the phasor diagram.

4. How does changing the values of R, L, and C affect the RLC series circuit?

Changing the values of R, L, and C in an RLC series circuit can affect several aspects of the circuit. Increasing the value of R will decrease the current and increase the power factor, while increasing the values of L and C will cause the circuit to become more inductive or capacitive, respectively. This will affect the impedance and phase difference between voltage and current, and ultimately, the power factor and efficiency of the circuit.

5. Can the RLC series circuit be solved using Kirchhoff's laws?

Yes, the RLC series circuit can be solved using both Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL). KVL states that the sum of voltages around a closed loop in a circuit is equal to zero, while KCL states that the sum of currents entering and leaving a node in a circuit is equal to zero. These laws can be used to set up equations and solve for the unknown values in an RLC series circuit.

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