Hello all, i need help figuring out my calculations. I was asked to calculate the capacitive reactance of a circuit with AC current. The capacitor has a value of 470nF. For 6 different Hz values (1,10,100,..,100k) i calculated the theoretical capacitive reactance. 1hz => 338799 ohms , 10hz => 33879,93 ohms and so on. I used 1 / 6.28 * Hz * 470 * 10 ^ -9

Now i made the circuit in the lab, calculated the Vc of the capacitor and the Ic and did Xc ( measured ) = Vc/Ic =>
For 1hz =>(too low of a frequency for a calculation) ,10 hz => 12.5 ohms, 100hz => 3.04 ohms and so on

Did i so something wrong? If not, what can i attribute to these staggering differences?

 

hi 40.
Welcome to AAC.
What is the output impedance of your 100Hz signal generator.?

E

A drawing showing your generator and voltage/current point measurements would help.

 

 

hi,
What is the total impedance between Node a and b.??

E

 

well, isnt it the impedance of the resistor + the impedance of the capacitor?

 

well, isnt it the impedance of the resistor + the impedance of the capacitor?

Hi,
In your post #1, you have only calculated the capacitive impedance, but in your actual test there is a resistor in series with that Xc value.
E

 

so i should remove the value of the resistor from the Theoretical calculations? Or should i make the remark that in the theoretical calculations, we do not account for any value of any resistor?

 

Hello all, i need help figuring out my calculations. I was asked to calculate the capacitive reactance of a circuit with AC current. The capacitor has a value of 470nF. For 6 different Hz values (1,10,100,..,100k) i calculated the theoretical capacitive reactance. 1hz => 338799 ohms , 10hz => 33879,93 ohms and so on. I used 1 / 6.28 * Hz * 470 * 10 ^ -9

Now i made the circuit in the lab, calculated the Vc of the capacitor and the Ic and did Xc ( measured ) = Vc/Ic =>
For 1hz =>(too low of a frequency for a calculation) ,10 hz => 12.5 ohms, 100hz => 3.04 ohms and so on

Did i so something wrong? If not, what can i attribute to these staggering differences?

How did you measure the current? If you got 12.5Ω you must have measured almost 1 Amp.

 

View attachment 295933

Always ask if the answer makes sense.

You've done that -- that's why you came here.

If the answer doesn't make sense, ask if the data upon which it is based makes sense.

Your current must be going through R1, right?

What must the voltage across R1 be if you have 0.08 A flowing through it? What about when you have 0.43 A flowing through it?

Does that make any sense at all?

 

How did you measure the current? If you got 12.5Ω you must have measured almost 1 Amp.

With a 1 Vrms source?

 

Using LTSPICE with the values you show and 100Hz, it get:

Vc = 0.8V Ic = 250uV

The equivalent resistance is 0.8 / 0.00025 = 3200Ω

But note that that calculation is not correct, because the current and voltage are out of phase. You cannot treat a capacitor under AC like a resistor, you need to do a vector calculation to get impedance, which is a complex number. It this case, it is not far off from the scalar math.

 

Using LTSPICE with the values you show and 100Hz, it get:

Vc = 0.8V Ic = 250uV

The equivalent resistance is 0.8 / 0.00025 = 3200Ω

But note that that calculation is not correct, because the current and voltage are out of phase. You cannot treat a capacitor under AC like a resistor, you need to do a vector calculation to get impedance, which is a complex number. It this case, it is not far off from the scalar math.

Unless the capacitor in your simulation has a resistive component, the only thing incorrect in that calculation is that it is not the equivalent resistance, but rather the reactance that is being calculated and the reactance is -3200 Ω. The impedance would simply turn out to be -j3200 Ω.

But the TS isn't struggling with the math or with a simulation. Rather, with data taken in the lab -- and there is strong evidence that the problem is that the data is simply wrong. There might be an issue with the set up, or with the instrumentation, or with interpreting what the instrumentation is reading.

 

well, isnt it the impedance of the resistor + the impedance of the capacitor?

Yes, this statement is correct, strictly speaking. However, I don't think it means what you probably think it means.

The "impedance" of a resistor is a real number, while the "impedance" of a capacitor is an imaginary number. This has to be taken into account when they are added together -- meaning that they they have to be added as complex numbers (or vectors, take your pick).

But, according to your set up, that is not a factor if you are trying to find Xc.

Presumably you are using meter XMM5 to measure the current. Assuming (and we'll get to this assumption later) that the meter has very low impedance and that the two meters being used as voltmeters, XMM6 and XMM7, plus the scope have very high impedance inputs, then XMM5 is measuring the current in a series circuit and the meter, source, resistor, and capacitor all have that same current, which you appear to be calling Ic.

Similarly, the voltage across the capacitor, which you appear to be calling Vc, is being read by XMM6 and, assuming it has sufficiently high impedance, should not be affecting the measurement of that voltage.

What you don't know is what the phase relationship is between the current in the capacitor and the voltage across it. Without that information, you have to make the assumption that the capacitor is a pure capacitance such that the current leads the voltage by 90°. Under that assumption, the reactance is then -Vc/Ic.

You can remove the assumption about the phase relationship by measuring it using the scope.

While modern real voltmeters, even cheap ones, tend to have very high input impedances (typically 10 MΩ), real ammeters, even some pretty expensive ones, often don't have particularly low input impedances. But I would expect them to be pretty low relative to 2.2 kΩ unless it is set on a very low range. Having said that, with a 1 Vrms source and capacitive reactances in the -34 kΩ range, you are trying to measure currents in the 30 µA range.

A generally better way to measure currents is to measure the voltage across a resistor that is in series with the current you want to measure. Normally, you need to make this resistor as low-valued as you can work with, but in this case you have a fixed, series resistor that is part of the circuit, namely the 2.2 kΩ resistor. So measure the voltage across that resistor and determine the current from that. Even with and Xc of -34 kΩ the voltage across that resistor will be about 66 mV and most meters have a 200 mV range, which is well-suited for this.

Another thing to consider is whether your meter supports, and is set up, to measure AC voltage and AC current -- and also how it measures those things, though if your waveforms are good approximations of sinusoids with no DC offset, that is less of a concern. But what is very possibly a concern is the frequency range over which the AC measurements are valid. If you are outside that range, the measurements you get are going to be off, perhaps WAY off.

To help us help you better, how, exactly, are you actually making your measurements? What meters are you using (make and model)? What function are they set on? What range are they set on?

But your best instrument to make these measurements (or, rather, the best instrument you are likely to have access to) is that oscilloscope. What does IT say?

 

so i should remove the value of the resistor from the Theoretical calculations? Or should i make the remark that in the theoretical calculations, we do not account for any value of any resistor?

Where are you using the value of the resistor in your theoretical calculations to begin with?

 

wow first of all thanks for all the replies you are very kind
the measurements i took at the lab were approved by the professor
either way the assignment has been submitted.

Where are you using the value of the resistor in your theoretical calculations to begin with?

in my assignment I said nowhere

 

I think that the reading are mA not Amps.
For 100Hz the total impedance would be 4038Ω, which would give a current of 247uA from a 1V source.
Similarly, the current at 1kHz would be 449uA, which agrees with the table assuming the units are wrong.

 

the measurements i took at the lab were approved by the professor

That a professor would approve measurements that claim that claim hundreds of milliamps are flowing through a 2.2 kΩ resistor from a 1 V source is troubling, but sadly not all that surprising.