# How can we test relationship between frequency and peak current of an alternating current?

It has been suggested to me that $$\frac{V_{peak}^2}{I_{peak}^2}=R^2_{coil}+4\pi^2f^2L^2$$ where $L$ is a constant and $f$ is the frequency of AC.

As a high school student I don't understand how can we keep $V_{peak}$ constant, vary $I_{peak}$, and measure change in $f$. I just simply don't understand anything. I have never seen an AC power supply or a signal generator. I have always encountered DC circuits. Is there anyway I can test this relationship in a laboratory with a coil, CRO, signal generator, multimeter and other commonly used equipment. I don't even understand what is need of a coil.

n.b. Originally the equation I was given was indeed wrong and that was what had to be shown in the lab but I am writing the correct equation to avoid any confusion.

• There is a conceptual problem here: Your circuit will determine the frequency $f$ at which it produces/sees/or does whatever with an AC voltage and current---and if your formula accurately describes it in the first place. So altering it (and be it externally) to force the frequency to adapt may not work as expected because this formula may turn out to be less fundamental than the reality given by the altered overall circuit! – pyramids Apr 5 '15 at 22:39
• @pyramids Your comment is possibly wrong or possibly just coming through very badly, but a driven circuit will always run at the driving frequency and not at it's natural frequency (a fact that always surprises me and is very important in a number of applications). This question is presumably getting at the resonant behavior which tells us that the size of the signal in the circuit is dependent on the how closely the driving frequency matches the natural frequency. – dmckee Apr 5 '15 at 23:21
• Suchal, it is really just a matter of picking the right tools. A good oscilloscope will allow you to simply measure $V_{peak}$, $I_{peak}$ and $f$. – dmckee Apr 5 '15 at 23:23
• @dmckee, to be sure, this is a 1st order problem (an inductor with internal resistance modelled) so is there resonance? (Yes, physical inductors have non-zero parasitic capacitance and thus, a self-resonance frequency but we're not modelling that here). As far as I can tell, this is simply (the magnitude of the) inductive reactance in series with internal resistance. – Alfred Centauri Apr 6 '15 at 1:38
• @AlfredCentauri Hmmm ... yes, I see. That said, there shouldn't be a problem in treating it that way as long as you avoid going to DC (where the various divergences will need to be handled with L'Hopital's rule. – dmckee Apr 6 '15 at 1:51

The question and some of the comments overlook the significance of the coil which has both a resistance R an inductive reactance (2 pi f L). Basic AC theory explains that a coil has inductive reactance "resists" changes in current such as continually occur when there is an alternating voltage supply, but has no effect on a direct current. The direct current case will be seen by making f=0 whereupon the equation reduces to simple Ohm's Law. However for alternating frequencies, the term (2 pi f L) has a value which is also in ohms but 90 degrees phase advanced relative to R ohms. We need proper AC theory to apply Ohm's Law here and should realize that the suggested equation is wrong; the left side should be written (Vpeak/Ipeak) squared. (Equivalently, take square root of the right side, as Alfred Centauri correctly says.)

To perform the measurement, one would use an AC source such as a Variac transformer (where f is the mains supply frequency either 50 or 60 Hz) or a signal generator (typically for audio frequencies 20 to 15 000 Hz). The source should deliver a sine wave and will likely be calibrated in Vrms where Vrms = 0.707 x Vpeak. You can use a multimeter to measure Vrms (N.B. ordinary multimeters measure r.m.s. AC voltage by assuming the wave is sine) or for better certainty you can use a CRO to measure on its calibrated graticule the peak-to-peak voltage (double Vpeak) of the sinewave.

Rcoil is easy to measure using the multimeter to measure ohms, this measurement is done with direct current. Measuring the current in the AC circuit is not as simple as measuring the voltage; you might add a relatively small resistor in series with the coil and measure the voltage drop across it (the extra resistor introduces some error, though, and should be considerably smaller than R).

The requested circuit diagram would simply show an AC supply connected to a coil where the coil is a realistic component that has a finite series resistance (as do all real coils not made superconducting). Comments that mention resonance and natural frequency are irrelevant as long as no capacitor is in the circuit.

• Well the equation is wrong and that is what we had to test in the lab. I had to design an experiment but I failed because I overlooked the importance of coil as you said. I would edit the question and write the correct equation to help anyone in the future. Someone commented that "AC ammeter" would be used in this case. I have never seen such ammeter. Does it exist? And why not measure p.d across the coil and the find current from I=V/R? I believe the resistance would be constant for all frequencies. – Suchal Apr 6 '15 at 20:08
• @Suchal, I thought that you understood that the formula only holds for an inductor (coil); this isn't a general result. It follows from the properties of inductance and the resistance of the conductor the coil is made from. For example, there would be different relationship between the AC voltage and current for a capacitor. – Alfred Centauri Apr 6 '15 at 21:45
• @alref yes I do. But is there anything like AC ammeter? – Suchal Apr 7 '15 at 15:24
• – Alfred Centauri Apr 7 '15 at 21:40

First, the formula as written isn't dimensionally consistent and so, cannot be correct. The correct formula$^1$ is

$$\frac{V_p}{I_p}= \sqrt{R^2_{coil} + (2\pi f L)^2 }$$

Now, this formula assumes sinusoidal (AC) voltage and current. Given a frequency $f$, the peak voltage across and peak current through are related by this formula.

Ordinarily, to test this relationship, one would connect the coil (with an AC ammeter in series and an AC voltmeter across) by connecting a signal generator, setting the frequency to some appropriate value, and then measuring the voltage and current.

This is then repeated at several different frequencies and the results are then compared with the theoretical formula.

$^1$I interpret the context as an otherwise ideal inductor with non-zero 'winding' resistance. In that case, the phasor voltage and current are related by

$$\frac{V_l}{I_l} = R_{coil} + j\omega L = R_{coil} + j2\pi f L$$

Taking the magnitude of both sides yields

$$\frac{|V_l|}{|I_l|} = \frac{V_p}{I_p} = |R_{coil} + j2\pi f L| = \sqrt{R^2_{coil} + (2\pi f L)^2 }$$

• A circuit diagram would be appreciated. I don't understand why there is a need of a coil? – Suchal Apr 6 '15 at 13:26
• – Alfred Centauri Apr 6 '15 at 13:28
• I meant to say what is role of coil in this case. Can't a signal generator generate AC current? – Suchal Apr 6 '15 at 13:34
• @Suchal, for there to be signal generator current, there must be something connected to the signal generator output terminals through which the current circulates. With nothing connected to the generator output terminals, there is voltage across but zero current. – Alfred Centauri Apr 6 '15 at 14:04
• Sir I request you to please draw a circuit diagram. If I were to do it then I would connect a CRO in parallel with a variable frequency power supply and ammeter in series with a resistor and then vary frequency and measure f using CRO and I using ammeter. I don't see need of a coil here. What is the purpose of the coil? – Suchal Apr 6 '15 at 15:30