How to get phase shift in this task and in general?

Question

So the task says:

In the serial circuit connected to the 50 Hz frequency alternating voltage, effective voltage values 𝑈 = 220 V, known voltages 𝑈L = 660 V and 𝑈C = 500 V. The current in the circuit is 11 A. Determine the 𝑅, 𝐿 and 𝐶, and phase shift φ between voltage and current

This is how circuit looks:

I have calculated Xl , Xc R L and by applying ohm's rule R = U / I = 220 / 11 = 20 Ω. Xl = UL / I = 660 / 11 = 60 Ω , Xc = Uc / i = 500 / 11 = 45.45 Ω , and then to find L and C I used formula Xl = wL , where w represents angular frequency, w = 2 * pi * f = 100 pi rad/sec, and the same for C, Xc = 1/ wC. And to find phase shift I used formula Φ = tg-1(X / R) = tg-1( ( 60 - 45.45) / 20) = 36.04 °

Can someone help?

Answer 1

Hint: First take the phase of the current signal as 0.

Then you will know the phase of the voltage signals across each element (R, L, and C) due to the nature of those components.

Knowing the phase and voltage (because they're givens in the problem) across L and C, you can find the values of L and C.

Now you only have to find a resistor value that gives 220 V magnitude across the source.

From there you'll get the phase of the source.

Then, since the problem asked for the phase of the current relative to the voltage source rather than the phase of the voltage source relative to the current, you'll just negate the phase and report that as the current phase. (I.e. instead of saying "the voltage leads the current by x radians" you'll say "the current lags the voltage by x radians")

Answer 2

Well, the RMS-voltages across the coil and capacitor are given by:

• Current: $$\text{I}_{\text{i},\space\text{rms}}=\text{V}_{\text{i},\space\text{rms}}\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)}}\tag1$$
• Coil: $$\text{V}_{\text{C},\space\text{rms}}=\text{V}_{\text{i},\space\text{rms}}\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)}}\cdot\frac{1}{\omega\text{C}}\tag2$$
• Capacitor: $$\text{V}_{\text{L},\space\text{rms}}=\text{V}_{\text{i},\space\text{rms}}\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)}}\cdot\omega\text{L}\tag3$$

Using your values we get:

$$\text{C}=\frac{1}{6000 \pi }\approx0.000053\space\text{F}\space\wedge\space\text{L}=\frac{3}{5 \pi }\approx0.19\space\text{H}\space\wedge\space\text{R}=10 \sqrt{2}\approx14.14\space\Omega\tag4$$

I used the following Mathematica-code:

In[1]:=w = 2*Pi*50;
FullSimplify[
 Solve[{11 == (220)*(1/Sqrt[2])*(1/Sqrt[R^2 + (w*L - (1/(w*c)))^2]), 
   660 == (220)*(1/Sqrt[2])*(1/
       Sqrt[R^2 + (w*L - (1/(w*c)))^2])*(1/(w*c)), 
   660 == (220)*(1/Sqrt[2])*(1/Sqrt[R^2 + (w*L - (1/(w*c)))^2])*w*L, 
   c > 0 && R > 0 && L > 0}, {c, L, R}]]

Out[1]={{c -> 1/(6000 \[Pi]), L -> 3/(5 \[Pi]), R -> 10 Sqrt[2]}}

In[2]:=N[%1]

Out[2]={{c -> 0.0000530516, L -> 0.190986, R -> 14.1421}}

And for the phase difference we can look at:

$$\arg\left(\underline{\text{V}}_\text{i}\right)-\arg\left(\underline{\text{I}}_\text{i}\right)=\arg\left(\underline{\text{Z}}_\text{i}\right)=\arg\left(10\sqrt{2}+60\text{j}-60\text{j}\right)=0\tag5$$