# Help with RC Circuit

Discussion in 'Homework Help' started by cts_casemod, May 14, 2013.

1. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
Hi All,

I come across this problem for my assignment which is a bit different from the ones I used to deal with,

I have a 130uF Capacitor in series with the mains (230V @ 50Hz) to feed a 32Ohm resistor with a voltage 160V and a current of 5Amps.

This circuit has a PFC of 44 and draws 7.8Amps from the mains because of the reactive power. I am asked to find a way to correct the power factor, so that my load is now only 3.5Amps.

How can I solve this problem?
Many Thanks

2. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
What on earth does it mean to have a PFC of 44?

You need to show your best attempt at a solution, and we can then use that as a springboard for guiding you along.

3. ### jcb19 New Member

Feb 20, 2013
14
0
If PFC is referring to power factor, then this is very wrong as Power factor should be somewhere between 0 and 1 and if the current is lagging the voltage then the power factor is also said to be lagging. If the current is leading the voltage (i.e. the circuit is capacitive), then the power factor is said to be leading.

4. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
Hi,

Yes I am referring to a leading power Factor of 0.44 and I am asked to show this as a percentage, hence the 44%. Apologies for any confusion

I understand that I can add an inductor so that XL - XC = 0 and I would have a PFC of 1 as the resistor is linear and the current will be given by I = U/R , but that would short circuit the power source to the resistor and I am asked to maintain the 160V in the resistor. What am I missing?

Thanks

Last edited: May 14, 2013
5. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
Hence a PF of 44%, not a PF of 44. There is a huge difference (namely two orders of magnitude!).

You want to think in terms of adding something across the load to shuttle power energy back and forth between the two reactive elements while still dropping voltage across the capacitor.

6. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
Yes, apologies for that one!

So what are you trying to say, that even tough XC and XL cancel each other I still have the desired effect dropping the voltage to my load? Or did I got it wrong?

Thanks

7. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
Yes, because the L and C are still dropping voltage across them, but you put the load in parallel with one of them so that it sees the voltage dropped just across one of them.

Now, whether you can use a single component to get a PF of 1 without affecting the votlage across the load or not, my guess is that you can, but you'll just have to run the design and see.

8. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
So what you say is I create a voltage divider with the capacitor and the inductor sort of thing in which the capacitor and the inductor are still in series, but with the resistor in parallel to one of them right? So wouldn't this mean that I have some kind of losses as if I was using a resistive divider? Or will this just provide a path for reactive power?

Also, how would this depend on R, say if the value of R was changed to 100Ohm?

When you say run the design do you mean using a simulator? If so can you recommend any?

Sorry I am a bit confused, looks like I imagined this to work in a completely different idea from the beginning. Many thanks for the help!

9. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
Let's take a step back and start piecing the solution together one step at a time.

230V AC source at 50Hz and 0 phase

A 130μF capacitor in series with a 32Ω resistor.

Q1) What is the impedance of the capacitor?

Q2) What is the total impedance seen by the source?

Q3) What is the current produced by the source?

Q4) What is the voltage across the resistor?

Q5) What is the power factor for the circuit?

Q6) Does the calculator power factor agree with the stated power factor?

Calculate each of these, including the phase angles where appropriate. This will give us a good reference point to talk about how to modify the circuit from here.

10. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
Q1) The impedance of the capacitor is approximately 24Ohms
Q2) I can only calculate the total impedance seen by the source as 24Ohms in series with 33Ohms, so a total of 57Ohms
Q3) The current produced by the source is 7.4Amp @ 250V (As measured by the KW meter)
Q4) The Voltage across the 33Ohm resistor is 160V
Q5) The power factor for the circuit is 0.42@245V and 0.40 @ 250V (As measured by the KW meter)

I am not sure what you require on question 6.
My Total Input power at 250VAC is 795Watt, so given that I=P/U the real current is 3.18A and the total current provided by the source is 7.35Amp, so this gives me a calculated PF of 0.43

Last edited: May 14, 2013
11. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
Okay, this has revealed some of the problems you are having.

Impedance is a vector quantity and is represented either by a phasor or by a complex number. Both are equivalent and I don't know which way you are familiar with.

The impedance of the capacitor is -j24.5Ω or 24.5Ω at an angle of -90°.

This is important because when you add to impedances in series, the impedances add, just like you are used to with resistors, but they do so as vectors.

So the total impedance seen by the source is (32-j24.5)Ω or, adding 32Ω at an angle of 0° to 24.5Ω at an angle of -90°, 40.3Ω at an angle of -37.4°.

Now ask if the current of 7.4A makes sense. What would the current be if the capacitor were replaced by a short and you just had the 32Ω resistor there. Or, if the two elements are in series, how can the resistor have 5A and the voltage source have 7.4A?

12. ### WBahn Moderator

Mar 31, 2012
23,552
7,207
Where are you getting these numbers from? All you have stated is a particular capacitor in series with a particular resistor in series with a particular source. How are you determining these power, current, and poer factor measurements? Please provide a diagram of what you are doing.

13. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
These are part of a traction application I am designing.

The 32Ohm resistor is a DC voltage source (battery)

The 130uF Capacitor limits the current between the main voltage source (AC Mains) and the given battery and its in series with it.

A Full wave bridge rectifier converts the 160V AC output from the capacitor to the DC Voltage source.

The circuit has been working for the last 6 months, as a prototype at 320V. Since the goal was to build a 750V Inverter I am required to double the available charging power and I am not able to do so because of the poor power factor.

As with the values,
The PF, input power and input current are read with a KW meter.
The output current is read with an amp meter at the output of the bridge rectifier.

The AC cables get warm and there is quite a large inductive load when powering the circuit. My guess is that the capacitor is acting as a resistor, but at the same time returning the unused power back to the grid at a different phase angle, hence the excess current.

It makes sense from the point of view that I = U/R, so I = 250/33 = 7.57Amp.
In Fact if I attach the capacitor alone to the Power meter it actually reads 10.27Amps current with a power consumption of 10Watt, since the remaining energy is being returned to the grid.

Using same maths this gives me R = U/I -> R = 250/10.27 = 24Ohm
Using the XC Formula for C = 100uF and F = 50Hz returns 24.48Ohms, so both the power meter and the capacitor values seem correct.

Now, going around the mesh and subtracting the 160V Supply from the 250V Input I get 90V. Since I = U/R -> I = 90/24 I get I = 3.75 and this is the current I should also be seeing on the DC side.

Why the real power is twice I have no idea.
I am sure there is something really simple here, so what am I missing!?

14. ### cts_casemod Thread Starter Member

May 14, 2013
35
0
The problem of the mysterious currents on the mesh is solved.

The capacitor would charge when the voltage was over 160V and would retain this charge not able to return it to the grid when the voltage started to decrease since the diodes were not conducting. On the negative side the same would happen but when the voltage got above 160V the diodes started to conduct and instead of gradually charging the capacitor the previously stored voltage (90V) was suddenly released into the load. The capacitor would now start to charge again and the same would happen on the next cycle.

So part of the energy that should be returned to the grid as reactive power was send to the battery, explaining the excess current.

Having this problem solved I now have current proportional to the difference between supplies over the value of XC for the capacitor.

To Solve I added another capacitor in a voltage divider fashion. This capacitor discharges the first when the diodes are not conducting returning the reactive power back to the grid.

Power factor dropped a bit, so I am back trying to find a way to solve this.