# Phase difference RLC circuit

1. Jan 6, 2012

### Gavroy

hi

i have a textbooks that says that if you have a driven RLC at its resonance frequency and now you increase the inductance, then you would measure a phase difference between the external emf and the voltage at the capacitor that is now greater than pi/2

unfortunately, i do not know how to calculate(<-that would be the best thing) or explain this fact?
(sorry about my english.)

Last edited: Jan 6, 2012
2. Jan 6, 2012

### Gordianus

Do you have a series or parallel circuit?

3. Jan 6, 2012

### Ken G

I will assume you mean a series RLC circuit (the results can be transformed into the parallel RLC circuit, but that seems a less common application). The series driven RLC circuit has the exact same mathematics as a driven damped harmonic oscillator. The sinusoidal driving voltage acts like a sinusoidal external driving force on the oscillator, so in the analogy, voltages are like forces. That means the resistance, which produces a voltage drop of R*dQ/dt, is acting like the damping force in the oscillator analogy, so Q is acting like the "displacement" in the analogy. The voltage drop across the capacitor is Q/C, so C is acting like the inverse of the "spring constant." Finally, the voltage drop across the inductor is Ld2Q/dt2, so L acts like the "mass."

So now that we have reduced the series RLC circuit to a driven damped harmonic oscillator, your question can be answered in that more familiar case. Your question is then about the phase relationship between the driving and the response (where by "response" I mean the displacement, or in this case, the displaced charge Q). The restoring force is always pi out of phase with the response, and the "inertial force" (here the inductive term) is always in phase with the response (imagine a particle on a spring moving in a circle-- that's the simplest case of simple harmonic oscillator response, and the restoring force is opposite the displacement, while the "centrifugal force" is along the displacement). But your question has to do with the phase relationship between the driving and the response, which depends on the frequency of the driving.

If the driving is at the resonant frequency, the phase of the driving is always pi/2 advanced ahead of the response. This is because at resonance, the restoring force completely balances the inertial force (the restoring force provides the acceleration, think again about a mass on a spring going in a circle), while the driving force is completely cancelled by the damping force (once we reach time steady behavior). The damping force lags the displacement by -pi/2 phase, because drag points opposite to the velocity, so the driving must lead the displacement by pi/2.

If you now increase the driving frequency, then you are moving closer to the situation where the driving is at very high frequency. In that limit, the driving force is becoming closer to balancing the inertial force (the driving is the main force, it provides the acceleration and the other forces become more and more negligible at high frequency and low amplitude). If the driving balances the inertial force, then the driving has to be opposite to the diplacement-- now the driving is providing the "centripetal acceleration" (in the case of the mass on the spring going in a circle again). So moving toward this limit must be moving the phase of the driving more in advance of the displacement, to something larger than pi/2 (and in the limit of huge frequency, all the way to pi).

Note that the mass on a spring going in a circle is always the easiest case to picture, and the connection to arbitrary oscillators (and RLC circuits) is made by the fact that sinusoidal motion in one dimension (say the dimension of Q) can always be obtained by superimposing two solutions that correspond to motion in a circle in the two opposite directions. Since the phase relationships I refer to above didn't depend on the direction of the circular motion, it all applies when you superimpose two circular motions going in opposite directions.

4. Jan 6, 2012

### technician

If it is a series circuit at resonance then Xl =Xc, Vl = Vc And the supply voltage is in phase with the current.
The voltage across the L will be 90 degrees ahead of the current (ahead of the supply voltage) and the voltage across the C will be 90 degrees behind the current (behind the supplyvoltage)
If L is increased then Xl will increase (Xl = 2πfL) This means that Vl will increase and be greater than Vc. The supply voltage will not now be in phase with the current, it will lead the current and therefore be greater than 90 degrees from Vc

5. Jan 6, 2012

### Ken G

And I should add that my answer imagined the driving frequency increased above resonance, whereas I see the original question instead had the inductance increasing for a fixed driving frequency. But that's the same effect, because increasing L reduces the resonant frequency, so either way the driving frequency is becoming larger than the resonant frequency. Also, I framed the phase relationship as that between the driving and the displaced Q at any given time, whereas the question asked about the phase between the driving voltage and the voltage at the inductor, but like I said, the voltage at the inductor is always in phase with displaced Q (in steady state), so again the answer is the same.

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