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Diagram RL This is a graph of voltages of output signal, of resistor, and of inductor, respectively. The cycle is just as the following:

┌─── R(100 Ohm) ─ L(8.2 mH) ───┐ └──────signal (5Hz sine)───────┘

Then the result shows that there is almost no difference of voltage phase between resistor and inductor, while the theoretical result says that there must be a difference of $\pi / 2$.

I guess there are some interferences by output signal, since inductance is very small($X_L = \omega L \approx 0.258~\mathrm \Omega$) comparing with $R = 100~\mathrm \Omega$; but I do not know why and how it interferes, or whether there are other affects.

Question. Why there is almost no phase difference between of resistor(2nd graph) and of inductor?(3rd graph)

p.s.: Every team has the same result, and other experiments with capacitor(with reactance $X_C\approx R$) give expected results whence I think it is not an experimental mistake.

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  • $\begingroup$ Have you done the analysis to see what phase difference you expect? $\endgroup$
    – The Photon
    Apr 29 '17 at 17:29
  • $\begingroup$ If you would please draw a proper circuit diagram, labeling everything including the points at which the voltages are measured, make sure it's obvious which voltages are which on the plots (right now I can't tell what ChA and ChB mean), then I will answer this question. Otherwise, this is a very unclear. I hope other users will refrain from giving an answer until this happens. $\endgroup$
    – DanielSank
    Apr 29 '17 at 17:30
  • $\begingroup$ @ThePhoton I think the phase difference must be 90 degs between $V_R$(resistor, 2nd graph) and $V_L$(inductor, 3rd graph), since $V_R \propto I$ and $V_L \propto \frac{\mathrm dI}{\mathrm dt}.$ $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 17:31
  • $\begingroup$ @DanielSank I added the circuit diagram; and I wrote in the question and the first sentence that the channel A(2nd graph) is for $V_R$ and B(3rd) for $V_I$. $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 17:41
  • $\begingroup$ Next question, do you know the coil resistance of your inductor? $\endgroup$
    – The Photon
    Apr 29 '17 at 17:52
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Here's my guess: the resistance of the wire making up the inductor is large compared to the inductive impedance at 5 Hz.

According to my calculations, the amplitude of the voltage across the inductor should be about 2.6 mV but the scope is reading about 50 mV.

Calculations:

At 5 Hz, the impedance of an ideal 8.2 mH inductor is given by

$$Z_L = j2\pi f L = j\, 0.082\, \pi\, \mathrm{\Omega} \approx j\,258 \mathrm{m\Omega} $$

Using phasor voltage addition, the phasor voltage across the inductor should be

$$1\mathrm{V} \frac{Z_L}{R + Z_L} = 1\mathrm{V} \frac{j0.258}{100 + j0.258} \approx 2.58 \mathrm{mV} \angle89.9^\circ$$

Now, stipulate that the inductor also has a resistance of 5 ohms. The phasor voltage across the inductor becomes

$$1\mathrm{V} \frac{Z_L}{R + Z_L} = 1\mathrm{V} \frac{5+j0.258}{100 + 5+j0.258} \approx 47.7 \mathrm{mV} \angle2.81^\circ$$

which is close to your experimental result.

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  • $\begingroup$ Thanks for your answer. I missed that fact. In fact, the values of $V$, $R$, $L$, $f$ are given by TA whence I cannot change them. $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 17:58
  • $\begingroup$ Yes, the fact that the inductor voltage is 50 mV is a pretty clear indication that the resistance of the inductor is a lot larger than it's reactance. $\endgroup$
    – DanielSank
    Apr 29 '17 at 18:01
  • $\begingroup$ A quick look at the Mouser catalog says 8.2mH inductors with a max current of 100mA (seems a reasonable guess, given a 5V supply and a 100R resistor) have resistances between 15R and 24R. A value of about 20R would be consistent with the OP's voltage measurements. $\endgroup$
    – alephzero
    Apr 29 '17 at 18:02
  • $\begingroup$ @alephzero I think the circuit was probabily this product; and there is a text 6.5 Ω maximum DC resistance in Specifications of inductor. Is this means $R_L \le 6.5~\mathrm{\Omega}$? If yes, I guess that the phase diff of $V_R$ and $V_L$ will be $$\phi = \tan^{-1}\frac{\omega L}{R_L} \ge 0.040~\mathrm{rad} \approx 2.27^\circ,$$ but the result is absolutely less than what I expected. What point I misunderstood? $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 18:20
  • $\begingroup$ @HalHollis Could you explain your calculation? $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 18:29
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New answer as a result of looking at some of the specifications supplied by PASCO.

PASCO state that their inductor has an inductance of 8.2 mH at 1 kHz and has a maximum DC resistance of $6.5\, \Omega$.
So assume that the resistance of the inductor is $6.5\, \Omega$.

At $5\,\rm Hz$ an $8.2 \,\rm mH$ inductor has a reactance of $2\times \pi \times 5 \times 8.2 \times 10^{-3} \approx 0.26 \, \Omega$.

So the phasor diagram looks like this.

enter image description here

The voltage of the supply leads the voltage across the resistor by $\phi = \tan^{-1}\left ( \dfrac {0.26}{106.5}\right ) = 0.14 ^\circ$ and the voltage across the non-deal inductor leads the voltage across the resistor by $\theta = \tan^{-1}\left ( \dfrac {0.26}{6.5}\right ) = 2.3 ^\circ$.

So measurement of phase angles is rather tricky.

Changing the frequency to $1 \, \rm kHz$ will produce this phasor diagram.

enter image description here

This should meant that the phase angles can be measured with reasonable accuracy and from them a better estimate of the resistance of the inductor might be found.


My original answer suggested a possibility of a problem with a common ground on each of the outputs and inputs.
If the interface which was used was a PASCO 550 Universal Interface or similar then the grounding problem goes away because such a device uses a differential input.

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  • $\begingroup$ Is this circuit right?: link $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 18:38
  • $\begingroup$ +) Then cannot I know how big the resistance of inductor is at 5 Hz? Or is there another method? $\endgroup$
    – Kanu Kim
    Apr 29 '17 at 19:22
  • $\begingroup$ Like you, I was (at first) convinced that that the connection shown would 'short out' the inductor. However, it does appear that CHA plus CHB is roughly equal to CH Vo $\endgroup$
    – Hal Hollis
    Apr 29 '17 at 19:58
  • $\begingroup$ @KanuKim This was an experiment. You could have measured it. The resistance of the inductor doesn't depend on the frequency. If you measured the phase angle at say 1kHz, you could calculate it, as in Farcher's answer, but at 5Hz the errors in your phase measurements would be too big. $\endgroup$
    – alephzero
    Apr 29 '17 at 23:39

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