What is the Phase Angle in a Series L-C Circuit?

In summary, when a 0.400H inductor and a 6 micro-Farad capacitor are placed in a series circuit with a voltage source of 30.0V and an angular frequency of 250 rad/s, the resulting impedance is primarily capacitive and the voltage will lag the current by -90 degrees. This is because the inductive and capacitive reactances partially cancel each other out, with the capacitor's reactance being larger in this case.
  • #1
MrLiou168
14
0

Homework Statement


You have a 0.400H inductor and a 6 micro-Farad capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0V and an angular frequency of 250 rad/s. What is the phase angle (phi) of the source voltage with respect to current?


Homework Equations



tan(phi) = (V_L - V_C) / V_R = (I(X_L - X_C)) / (IR) = (X_L - X_C) / R


The Attempt at a Solution



First, with no resistor, all the above equations diverge. Also, I figured that since in an inductor, v leads i by pi/2, and in a capacitor, v lags i by pi/2, the answer may be zero, but Mastering Physics deemed it wrong.

Any help greatly appreciated!
 
Physics news on Phys.org
  • #2
MrLiou168 said:

Homework Statement


You have a 0.400H inductor and a 6 micro-Farad capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0V and an angular frequency of 250 rad/s. What is the phase angle (phi) of the source voltage with respect to current?


Homework Equations



tan(phi) = (V_L - V_C) / V_R = (I(X_L - X_C)) / (IR) = (X_L - X_C) / R


The Attempt at a Solution



First, with no resistor, all the above equations diverge. Also, I figured that since in an inductor, v leads i by pi/2, and in a capacitor, v lags i by pi/2, the answer may be zero, but Mastering Physics deemed it wrong.

Any help greatly appreciated!

The only impedances involved are reactive in nature (XL, XC). So the question becomes, which dominates the overall reactance of the circuit? What values have you calculated for XL and XC? What's their signed sum? Is the result "inductive" or "capacitive" nature?
 
  • #3
I calculated X_L = 100 and X_C = 667 and I guess the signed sum is >0. And since the capacitive reactance is larger, I assume that capacitance dominates the circuit. How does this lead to obtaining a value for R in order to get the phase angle?
 
  • #4
MrLiou168 said:
I calculated X_L = 100 and X_C = 667 and I guess the signed sum is >0. And since the capacitive reactance is larger, I assume that capacitance dominates the circuit. How does this lead to obtaining a value for R in order to get the phase angle?

When series reactances are "summed", you will see the expression "XL - XC". This is a consequence of the general concept of impedance which you will no doubt cover in your courses at some point. Impedance is like resistance but uses complex numbers. The reactive "resistance" ends up in the imaginary component of the impedance, where inductive values have positive imaginary terms and capacitive values have negative imaginary terms. In this case, the individual impedances for the inductor and capacitor would look like ZL = jXL, and ZC = -jXC.

Anyway, to make a long story short, your overall circuit will "look" like a capacitor with reactance 566.6 Ω (or impedance Z = -j566.6 Ω). With no resistance (R) in the circuit it just looks like a capacitor. How does the voltage vs current angle look for a capacitor?
 
  • #5
Thanks, but I still don't understand. There is no mention of imaginary numbers in my textbook, and no equation that involves angles other than the ones I listed initially.

Should I simply consider reactance = resistance = R = 566.6 ohms and use it in my 2rd equation above?
 
  • #6
MrLiou168 said:
Thanks, but I still don't understand. There is no mention of imaginary numbers in my textbook, and no equation that involves angles other than the ones I listed initially.
You'll get to complex impedances eventually, if not in this particular course then in some other follow-on. Never mind it for now.
Should I simply consider reactance = resistance = R = 566.6 ohms and use it in my 2rd equation above?
Nope. Reactance isn't really the same thing as resistance, despite having the same units. What you've got, without resistance in the circuit, is something that's going to "look like" either a capacitor or an inductor. Which one it "looks like" depends upon which of the reactances dominates. Don't try to use a formula that requires a resistance when none is present. Instead, consider the relationship of current and voltage in pure inductances and pure capacitances. Hint: There's only two choices of angle!
 
  • #7
OK I think I'm beginning to get it... since capacitance dominates, it won't cancel the effect of the inductance, but the circuit will "look" like a capacitor, so voltage will lag current by pi/2...?
 
  • #8
MrLiou168 said:
OK I think I'm beginning to get it... since capacitance dominates, it won't cancel the effect of the inductance, but the circuit will "look" like a capacitor, so voltage will lag current by pi/2...?

Yup. Actually, the effects of the reactances are to mutually cancel, with the "winner" being the one that's larger. Here the inductor manages to "cancel" only a portion of the capacitor's reactance.
 
  • #9
Thanks for your help, much appreciated.

As a side note, for those of you attempting this problem, Mastering Physics does not recognize the answer if you enter 270 degrees. Use NEGATIVE 90 degrees.
 

1. What is a L-C series circuit?

A L-C series circuit is a type of electrical circuit that consists of a capacitor (C) and an inductor (L) connected in series. It is also known as an LC resonant circuit or a tank circuit.

2. How does a L-C series circuit work?

In a L-C series circuit, the capacitor stores electrical energy in the form of an electric field, while the inductor stores energy in the form of a magnetic field. The two components interact with each other, causing the circuit to oscillate at a specific resonant frequency.

3. What is the resonant frequency of a L-C series circuit?

The resonant frequency of a L-C series circuit is determined by the values of the capacitor and inductor. It can be calculated using the formula f = 1 / (2π√(LC)), where f is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.

4. What are the applications of L-C series circuits?

L-C series circuits have various applications in electronic devices such as radios, televisions, and oscillators. They are also used in power factor correction, voltage regulation, and frequency filtering in electronic circuits.

5. What are the potential problems with L-C series circuits?

The main problem with L-C series circuits is that they can experience high voltage and current peaks, which can lead to damage or failure of the components. Additionally, the resonant frequency can be affected by external factors such as temperature and component tolerances.

Similar threads

Replies
3
Views
375
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
843
  • Introductory Physics Homework Help
Replies
10
Views
538
  • Introductory Physics Homework Help
Replies
15
Views
4K
Replies
17
Views
509
  • Introductory Physics Homework Help
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top