Find Effective Voltage Across R,L,C in RLC Circuit

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In summary, the effective voltage across each element (R,L,C) of a series RLC circuit is Vm/√2. The formulas for the peak voltages for the individual elements are VR = Im*R, VL = Im*XL, VC = Im*XC. The current lags the supply voltage so the supply voltage LEADS the voltage across R.
  • #1
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I am trying to find the effective voltage across each element (R,L,C) of a series RLC circuit like the one below. I am given the following values: Ieff = 1 Amp, Veff = 1 V, i(t) lags v(t) by 45°, L = 1, f = 2/2∏ Hz, L = 1 H
RLC_series_circuit.png


What I have done so far:
Veff = Vm/√2
=> Vm = √2
Ieff = Im/√2
=> Im = √2
XL = ωL = 2

v(t) = √2*sin(2t)
i(t) = √2*sin(2t - 45°)

i know the formulas for the peak voltages for the individual elements are
VR = Im*R
VL = Im*XL
VC = Im*XC

i'm not sure if those correspond to the effective voltages or if i have to use the equation Veff = Vm/√2
i am also having trouble finding R and XC from the information provided..
 
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  • #2
I agree with your value for Xl.
If the current lags the supply voltage then the supply voltage LEADS the voltage across R because the voltage across R and the current are in phase.
Are you familiar with showing the voltages Vl, Vr, and Vc on a vector (phasor) diagram. Vl is drawn on the =+y axis, Vr is drawn on the +X axis (horizontal axis ) and Vc is drawn on the -y axis.
This means, from the information in your question, that the supply voltage (1volt) will be a vector from the origin at 45 degrees to the X axis.
Hope this is of some help
I think that you have not been given enough information! Do you know anything about R or C?
 
  • #3
Sorry...you do have enough information,I had missed the current
 
  • #4
Imax = Vmax/Z
=> Z = 1
(1) Z = sqrt(R2 + (XL - XC)2)
cosθ = R/Z, θ = 45°
1/√2 = R/1
R = 1/√2
From (1) I get C = 1/(4-√2)

I am not 100% sure this is correct
 
  • #5
I got those values for R and C
 
  • #6
technician said:
I got those values for R and C

thanks for the help. i got
VR = Im*R = 1
VL = Im*XL = 2√2
VC = Im*XC = 2√2 - 1
 
  • #7
I got
Vr = 1 x Cos45 =0.7V
R = 0.7/1 = 0.7Ω
Vc = 1.3V
Xc = 1.3 (gives C= 0.38F)

I think the differences are due to the fact that I used the voltage and current values in the question. I assume these are rms values and I think you have converted these to peak values.

A good result !
Cheers
 

1. What is an RLC circuit?

An RLC circuit is a type of electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These three components are connected in either series or parallel and interact with each other to create a complex electrical response.

2. How do I find the effective voltage across an RLC circuit?

To find the effective voltage across an RLC circuit, you can use the formula V = IZ, where V is the voltage, I is the current, and Z is the impedance. The impedance of an RLC circuit can be calculated using the individual impedances of the resistor, inductor, and capacitor, which are determined by their respective values and the frequency of the circuit.

3. What is the significance of finding the effective voltage in an RLC circuit?

The effective voltage in an RLC circuit is important because it allows us to understand how the circuit responds to different frequencies. By finding the effective voltage, we can determine the amount of current and power that will flow through the circuit, and make adjustments to optimize its performance.

4. Can the effective voltage change in an RLC circuit?

Yes, the effective voltage in an RLC circuit can change depending on the frequency of the circuit. This is because the impedance of each component (resistor, inductor, and capacitor) varies with frequency, which in turn affects the overall impedance of the circuit and the effective voltage.

5. How can I use the effective voltage to troubleshoot problems in an RLC circuit?

If you are experiencing issues with an RLC circuit, you can use the effective voltage to troubleshoot the problem. By measuring the voltage at different points in the circuit, you can identify where the voltage drops or spikes, indicating potential issues with specific components. This can help you pinpoint the source of the problem and make necessary repairs or adjustments.

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