hi,

I have a L-R circuit in series which L=100mH and R =27ohm
My AC supply voltage is 12Vac,50Hz.

After i calculated the expected result,i get XL = 31.4ohm, Z = 41ohm, Is = 0.29A, Vr = 7.83V.

But, from what i measured from circuit i get this. I = 32.4mA,Vr = 0.78V.

Can someone help me on this?

thank you.

 

Correct me if I'm wrong, but....
Judging by your measured results. If the only resistor is 27ohms and the current is 32.4ma. You are looking at around 1200mH inductor to have that much of a voltage drop at 50Hz. From your measured result I get the following:
\(12v-0.78v=11.22V_{L}\)
\(11.22v/32.4ma=346.3ohms=X_{L}\)
\(Z_{total}=12v/32.4ma=370ohm \)


if \(a^2+b^2=c^2 \)

then \(27^2+b^2=370^2\)

\(b^2=370^2-27^2\)

\(b=\sqrt{370^2-27^2}\)

\(b=369\Omega\)

Almost there...

\(X_{L}=2\Pi fL\)

\(369=2\Pi50L\)

\(369/2\Pi50=L\)

\(L=1.175H \)

So maybe double check the size of that inductor.

NOTE: The actual calculated value for R would be 23.7Ω, but that doesn't change the rounded calculations in this case.

 

Correct me if I'm wrong, but....
Judging by your measured results. If the only resistor is 27ohms and the current is 32.4ma. You are looking at around 1200mH inductor to have that much of a voltage drop at 50Hz. From your measured result I get the following:
\(12v-0.78v=11.22V_{L}\)
\(11.22v/32.4ma=346.3ohms=X_{L}\)
\(Z_{total}=12v/32.4ma=370ohm \)

You can't just subtract 0.78V from 12V and expect a reasonable answer.
The calculation depends on phase shift.

The voltage across the inductor will be 90 degrees out of phase from that of the resistor.

Using the 27 ohms and 100 mH his calculation is correct;
Voltage across the resistor = 7.8 Vrms
Voltage across the inductor = 9.1 Vrms
Supply voltage = 12 Vrms

Since they are 90 degrees apart the relationship is actually:
\(V_{src} = \sqrt{V_R + V_L}\)

@simponss;
How exactly did you measure those values and with what devices?

 

hi Ghar,

i'm using a Digital multimeter which is in AC mode to measure all the current and voltage values. For Vr i put the probe across the two leg of the resistor. Any idea on it?

anyway thanks for the replies.

 

So for the current measurement you did use the ammeter?

Anyway, I do agree with c_omalley, the inductor may be much larger than you think.
There are more possibilities though, did you measure the voltage at the supply (with load attached) and across the inductor?

 

DOH! *slaps forehead*

@Ghar
Yeah, it's been a while...that's why I said "correct me if I'm wrong".
My error and apologies. I got the same answers as simpsonss with the stated values of 100mH, 27ohms, and 12vac, but I got in a hurry and made a SILLY mistake working backwards. I can't believe I did that. Thanks for pointing that out.
I apologize for posting before double-checking my work.

Since they are 90 degrees apart the relationship is actually:
\(V_{src} = \sqrt{V_R + V_L}\)

I believe you meant to type:
\(V_{src} = \sqrt{V_R^2 + V_L^2}\)

But I knew what you meant at any rate


* \(V_{src}^2=V_R^2+V_L^2}\)
* \(12Vac^2=0.78V_r^2+V_L^2\)
* \(144Vac=0.608V_r+V_L^2\)
* \(V_L=\sqrt{144-0.608}\)
* \(V_L=\sqrt{143.4}=11.97Vac_L\)

Is that correct?

@simpsons
Measure your supply output. It's possible the supply is overloaded. (i.e. rated for 200ma trying to drive a circuit requiring 300ma). Measure the inductor voltage drop. If your supply is putting out 12Vac and you only have 780mVac across the resistor, you should have a large voltage drop across the inductor. Around 11.9Vac. If not check all of your wires and connection. A faulty connection or wire will add resistance to you circuit.

Good Luck