# LRC-circuit

Discussion in 'Homework Help' started by kvi037, May 17, 2013.

1. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Hi,

A 60-cycle/sec source of emf is impressed across a series circuit consisting of a 10-ohm resistance and a capacitance of unknown value.
Rms voltage is 100 volts and rms current is 0.2 amperes.

Calculate
a) the impedance of the circuit,
b) the power consumption,
c) the power factor,
d) the value of capacitance

Last edited: May 18, 2013
2. ### #12 Expert

Nov 30, 2010
18,078
9,619
You should be putting these in the homework section.

3. ### maitchy New Member

Jan 22, 2010
7
1
This LCR circuit doesn't have an "L". Maybe the question should go in a Christmas homework section? (No L, No L, No L, No L. Born is the King...)

4. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Ok, I think we have to assume no XL-value in the circuit.

Can we start to say that Z = Urms/ Irms = 500ohms

That gives a too high XC-value compared to R.

Some suggestion on a better way to do it?

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,909
1,738
Does it? Show your calculation and your cross checking.

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,909
1,738
Why do you think that is incorrect?

7. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Z = ((R^2 -(XL -XC))^2 )^1/2

Assumes that XL is 0.

500= (100 - XC^2)^1/2

250000 = 100 - XC^2

XC = 500,1

8. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Bacause I haven't taken the power factor into account.

9. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,909
1,738
You haven't solved the power factor question yet ....

Why did you delete your post with the answer of 499.9 and change it to 500.1? You went from close answer, as you rounded it to one decimal place to an incorrect answer.

10. ### WBahn Moderator

Mar 31, 2012
23,087
6,940
Where do you get that you subtract off the reactance? Think of it in terms of the sides of a triangle -- the resistance is one side and the net reactance is the other. The impedance is the length of the hypotenuse, which is the square root of the sum of the squares of the sides.

And please start using units in your work.

11. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Understand, XC should be = 499.9 ohms

And for the power factor we have:

P = I^2 * R = 0.2^2 A *10 = 0.4 W
S = I^2 * Z = 0.2^2 A * (10+499.9) ohms = 20.4 VA

PF = 0.4/ (20.4) = 0.0196

12. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
But I haven't taken the 60-cycle/sec source into account, which making this calculation incorrect.

13. ### WBahn Moderator

Mar 31, 2012
23,087
6,940
You are at least trying to track units. Sort of.

Power consumption:

P = I^2 * R = (0.2A)^2 *10Ω = 0.4 W

A^2 times Ω is W. You can convert them to base (or at least more basic) units and verify this. You need to get in the habit not only of writing down the units, but tracking them. Meaning that when you write the 0.4W you ask yourself if amps squared time ohms is watts and only proceed if you are sure that it does.

Now, S is apparent power (or, more accurately, complex power with apparent power being the magnitude of it).

S = I^2 * Z = (0.2A)^2 * (10+499.9) ohms = 20.4 VA <== WRONG!

The magnitude of the impedance is NOT the algebraic sum of the resistance and the reactance. This is a vector quantity. But you also know that the apparent power is simply the rms voltage multiplied by the rms current. so what does that give you.

14. ### WBahn Moderator

Mar 31, 2012
23,087
6,940
Okay, so if you know that not taking the 60-cycle/sec into account somehow makes the calculation incorrect, why didn't you take it into account?

What difference would it have made in the result?

15. ### kvi037 Thread Starter New Member

May 17, 2013
17
0
Thanks for reply. I think I have figured out the solution.

a) 500 ohms

b) 0.2 ^2 A * 10 ohms = 0.4 W

c) S= Vrms * I rms = 20 VA

PF = P/S = 0.4 w / 20 VA = 0.02

d) PF gives a phase ange between voltage and current: 88.9

Xc = Z sin 88.9 = 499.9 ohms

16. ### WBahn Moderator

Mar 31, 2012
23,087
6,940
You've basically got it. There are just a couple of minor points you are still messing up.

0.2 ^2 A * 10 ohms = 0.4 W

You need to square 0.2A to get 0.04 A^2 and not 0.04 A. The units are physical quantities that take part in the same operations as their coefficients. If you don't do that correctly, then they are meaningless. If you do do it correctly, you can catch the majority of your mistakes which will result in better grades now and less death and distruction later. Remember, doctors kill people one at a time, engineers do it in job lots!

Second, the phase angle is -88.9 degrees. Strictly speaking, reactance is a signed quantity that is positive for inductive reactance and negative for capacitive reactance. Not all authors are careful about that, which is a shame, but if you are only going to treat all reactance as a magnitude only, then you have to explicitly track and state whether it is inductive or capacitive at every step, which is cumbersome and error prone. Far better just to have inductive reactance be positive and capacitive reactance be negative. Then you can add reactances and let the chips fall where they will and interpret the results at any given point based on whether the reactance is positive or negative.

kvi037 likes this.
17. ### WBahn Moderator

Mar 31, 2012
23,087
6,940
Oh, and you need to check to see if your answers actually answer the questions that were asked.

a) the impedance of the circuit,

This should consist of two parts, a magntitude and an phase angle.

b) the power consumption,

This should be in watts and is the power actually dissipated by the circuit.

c) the power factor,

This should be a number less than or equal to 1 and an indication of whether it is leading or lagging.

d) the value of capacitance

This should be in farads.