Solve RLC Series Circuit: V, f, L, C Given

In summary, in this RLC series circuit, the rms potential difference provided by the source is 150 V at a frequency of 220/pi Hz. Given the values of L = 0.7 H, C = 70 µF, and V_R = 80 V, the following can be determined: a) I (rms) = A b) R = Ω c) V_L (rms) = V d) V_C (rms) = V Using the equations X_L=WL_i and X_C=1/WC, where w= 2(pi)(f) and the voltage divider rule, the magnitude
  • #1
MarcL
170
2

Homework Statement



In an RLC series circuit, the rms potential difference provided by the source is V = 150 V, and the frequency is f = 220/pi Hz. Given that L = 0.7 H, C = 70 µF, and V_R = 80 V, find:

a) I (rms);
I = A

b) R;
R = Ω

c) V_L (rms);
V_L = V

d) V_C (rms).
V_C = V

Homework Equations



X_L=WL_i , X_C= 1/WC ; w= 2(pi)(f) ,

The Attempt at a Solution



A part from the equations listed above I don't really know what I need. I am kinda at lost with RLC circuits. my teacher taught it really quickly and it really confuses me. I know
I_rms = I (peak) / sqrt(2) but I'm not even sure how to get the peak current.

Any help to lead me in the right direction would be appreciated!
 
Physics news on Phys.org
  • #2
MarcL said:

Homework Statement



In an RLC series circuit, the rms potential difference provided by the source is V = 150 V, and the frequency is f = 220/pi Hz. Given that L = 0.7 H, C = 70 µF, and V_R = 80 V, find:

a) I (rms);
I = A

b) R;
R = Ω

c) V_L (rms);
V_L = V

d) V_C (rms).
V_C = V

Homework Equations



X_L=WL_i , X_C= 1/WC ; w= 2(pi)(f) ,

The Attempt at a Solution



A part from the equations listed above I don't really know what I need. I am kinda at lost with RLC circuits. my teacher taught it really quickly and it really confuses me. I know
I_rms = I (peak) / sqrt(2) but I'm not even sure how to get the peak current.

Any help to lead me in the right direction would be appreciated!

What about the basic circuit laws and methods? You know, like KVL, Ohm's Law, etc?
 
  • #3
Well I do know them, ( I don't know if you mean Kirchoff law by KVL?). I haven't been taught to apply them in RLC circuits. I know it sounds dumb because they never change ( i.e:voltage will be the same between 2 parallel conductor and so on). I am not very good with electricity, I try my best but it takes time to make sense in my head.

A part from that, how would I be able to apply these law for my current if the latter alternates?
 
  • #4
MarcL said:
Well I do know them, ( I don't know if you mean Kirchoff law by KVL?). I haven't been taught to apply them in RLC circuits. I know it sounds dumb because they never change ( i.e:voltage will be the same between 2 parallel conductor and so on). I am not very good with electricity, I try my best but it takes time to make sense in my head.

A part from that, how would I be able to apply these law for my current if the latter alternates?

Have you been taught about impedance (complex valued "resistance") yet, or only reactance for inductors and capacitors?
 
  • #5
I do know the formula for impedance, as in z= sqrt(R^2 (X_C+X_L)^2) but not the meaning of it or when to use it.
 
  • #6
MarcL said:
I do know the formula for impedance, as in z= sqrt(R^2 (X_C+X_L)^2) but not the meaning of it or when to use it.

Okay, that's actually a formula for the magnitude of the impedance of a series RLC connection, which will work for you just fine. Impedance is actually a complex-valued analog to resistance, but like any complex value it has a magnitude.

You need to be a bit careful about the signs you assign to the two reactances; in a series circuit capacitive reactance and inductive reactance work against each other when they are summed.

You can use that impedance magnitude like you would a total resistance to determine things like current magnitude. Hint: you might try using the voltage divider rule to find an expression for the voltage on the resistor (which you've been given a value for).
 
  • #7
Would it be weird if I said I haven't heard of the voltage divider rule? Plus, now if I understand, can I take the impedance magnetude ( in ohms) and use V = IR?? Or is that completely wrong??
 
  • #8
MarcL said:
Would it be weird if I said I haven't heard of the voltage divider rule? Plus, now if I understand, can I take the impedance magnetude ( in ohms) and use V = IR?? Or is that completely wrong??

Suppose you have a battery and two resistors in series (Say E, R1, and R2). Derive an expression for the voltage across R2. That will show you the form of the voltage
divider rule. It's very handy indeed and something well worth becoming familiar with.

Yes, Ohm's law applies for the magnitudes of the voltage, impedance, and current.
 
  • #9
ooooohh VDR would be R_2/R_T * V_T = V_R2! And can this rule be applied to find X_L and X_C?
Also, I have few more questions however... you said above

"You need to be a bit careful about the signs you assign to the two reactances; in a series circuit capacitive reactance and inductive reactance work against each other when they are summed. "

Does that mean in my impedance formula I'd have X_C-X_L?
 
  • #10
MarcL said:
ooooohh VDR would be R_2/R_T * V_T = V_R2! And can this rule be applied to find X_L and X_C?
It can. Although first you need to determine R (the resistor value). Your impedance magnitude expression serves as the "R_T" of the voltage divider expression in this case.

Also, I have few more questions however... you said above

"You need to be a bit careful about the signs you assign to the two reactances; in a series circuit capacitive reactance and inductive reactance work against each other when they are summed. "

Does that mean in my impedance formula I'd have X_C-X_L?
Yes. Or ##X_L - X_C##. Either way, since the result is squared so the sign of the difference doesn't matter in this case.
 
  • #11
And going back to the initial question, for an average i, I can use the average voltage but I don't have my resistance, even using the voltage divider rule I can't find it because I still have 2 unknown ( Z and R_3). I am stuck there because I tried to use V_rms = I_rms * Z
 
  • #12
You're given the voltage across the resistor. You're given the supply voltage. You have an expression for the total magnitude of the impedance which includes the R in it. Use the voltage divider rule to write an equation and solve for R; it doesn't involve I as a variable at all.
 
  • #13
Well I get 327 ohms for my R_3 by doing this:

R3 = (VR3/VT) √(R32+(XC+XL)2)

Which gave me R3 and to find my Irms = Vrms / Z which I got using my impedance formula, I got Irms = .33 A. It is still wrong though, I don't know if someone can tell me if its a calculation mistake or I got the wrong step?
 
  • #14
MarcL said:
Well I get 327 ohms for my R_3 by doing this:

R3 = (VR3/VT) √(R32+(XC+XL)2)

Which gave me R3 and to find my Irms = Vrms / Z which I got using my impedance formula, I got Irms = .33 A. It is still wrong though, I don't know if someone can tell me if its a calculation mistake or I got the wrong step?

I'm not sure how you arrived at that formula for ##R_3##, and again, beware of the sub-expression for the total reactance... the reactances don't just add. How did you derive the expression?

The results you're getting for the resistance and current don't look right to me.
 
  • #15
They aren't right ( according to my assignment).

I derived the formula for R3 from the voltage divider rule ( the one I derived earlier) where (R2/ RT) * VT=VR2

I just replaced RT with Z because my impedance is the total resistivity of the circuit ( unless I misunderstood). So because I have XL and XC in my circuit, respectively inductance reactance and capacitive reactance (I think those are the terms?), then I can find my Z but because I am lacking R3 I thought I could use the voltage divider rule to find my R3 as I had its respective voltage
 
  • #16
MarcL said:
They aren't right ( according to my assignment).

I derived the formula for R3 from the voltage divider rule ( the one I derived earlier) where (R2/ RT) * VT=VR2

I just replaced RT with Z because my impedance is the total resistivity of the circuit ( unless I misunderstood).
The term "resistivity" has a specific definition and doesn't fit here. Z is the magnitude of the impedance. No, you've understood correctly. But something has gone wrong with your implementation of the plan :smile:

So because I have XL and XC in my circuit, respectively inductance reactance and capacitive reactance (I think those are the terms?), then I can find my Z but because I am lacking R3 I thought I could use the voltage divider rule to find my R3 as I had its respective voltage

You don't have Z because your R3 is part of it and it is unknown. What you do have is the voltage across is though.

Suppose XL and XC are the reactances of the inductor and capacitor respectively, R is the unknown resistor, VR is the given voltage across VR, and E is the source voltage driving the circuit. Then the magnitude of the impedance of the series circuit is
$$Z = \sqrt{R^2 + (XL - XC)^2}$$
Note that R appears in that expression. Note also the use of the DIFFERENCE between the reactances; inductive and capacitive reactances CANCEL each other.

Use the above expression for Z as the total impedance for your voltage divider.
 
  • Like
Likes 1 person
  • #17
So I got it right but I used the ratio Vr / R3 = Vt/Z after realizing that it is proportional. I don't know if that's what you meant by voltage divider rule but it kinda made me think of this. Thank you!
 

1. How do I solve for voltage (V) in an RLC series circuit?

To solve for voltage in an RLC series circuit, you can use the formula V = IZ, where I is the current and Z is the total impedance of the circuit. The total impedance can be found by adding the individual impedances of the resistor (R), inductor (L), and capacitor (C) in the circuit.

2. What is the formula for frequency (f) in an RLC series circuit?

The formula for frequency in an RLC series circuit is f = 1 / (2π√(LC)), where L is the inductance and C is the capacitance of the circuit.

3. How do I find the inductance (L) in an RLC series circuit?

To find the inductance in an RLC series circuit, you can rearrange the formula for frequency to solve for L. It becomes L = 1 / (4π²f²C), where f is the frequency and C is the capacitance of the circuit.

4. Can I use the same formula to solve for capacitance (C) in an RLC series circuit?

Yes, you can use the same formula as in question #3 to solve for capacitance in an RLC series circuit. Simply rearrange the formula for frequency to solve for C: C = 1 / (4π²f²L).

5. What is the significance of solving for voltage, frequency, inductance, and capacitance in an RLC series circuit?

Solving for these values in an RLC series circuit allows you to analyze the behavior of the circuit and make predictions about its performance. It can also help you troubleshoot any issues in the circuit and determine the best components to use for optimal performance.

Similar threads

Replies
3
Views
394
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
17
Views
529
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top