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

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Homework Help Overview

The discussion revolves around determining the phase angle in a series L-C circuit consisting of a 0.400H inductor and a 6 micro-Farad capacitor, powered by a voltage source with a specified amplitude and angular frequency. Participants are exploring the relationships between the inductive and capacitive reactances and their effects on the phase angle of the source voltage with respect to the current.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of having no resistance in the circuit and how that affects the calculations for phase angle. They question the dominance of reactance types and how to interpret the results without using resistance.

Discussion Status

There is an ongoing exploration of the relationships between the reactances of the inductor and capacitor. Some participants have calculated specific values for the reactances and are considering how these values influence the overall behavior of the circuit. Guidance has been offered regarding the interpretation of the circuit's behavior in terms of capacitive and inductive characteristics.

Contextual Notes

Participants note that their textbooks do not cover complex numbers or imaginary components, which adds to the confusion regarding the calculations and interpretations of phase angles in this context.

MrLiou168
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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!
 
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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?
 
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?
 
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?
 
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?
 
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!
 
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...?
 
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.
 
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.
 

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