Calculating Capacitance and Inductance in an A.C. Circuit - What Went Wrong?

[chikis]

Homework Statement

A source of e.m.f 240V and frequency 50 HZ is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V.

Calculate the:
(i) potential difference across the capacitor,
(ii)capacitance of the capacitor,
(iii) inductance of the inductor

Homework Equations

The Attempt at a Solution

I got the answer to (i)
as 245V

that of (ii) I got to this extent C= 1/2450(pi)
= 1/7700
= 0.000129870
aprox. 0.0001299
= 1299*10^-4 uf

my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
What really happened?

 

[ehild]

Homework Statement

A source of e.m.f 240V and frequency 50 HZ is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V.

Calculate the:
(i) potential difference across the capacitor,
(ii)capacitance of the capacitor,
(iii) inductance of the inductor

Homework Equations

The Attempt at a Solution

I got the answer to (i)
as 245V

that of (ii) I got to this extent C= 1/2450(pi)
= 1/7700
= 0.000129870
aprox. 0.0001299
= 1299*10^-4 uf

my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
What really happened?

C = 0.0001299 F is correct. 1 F = 10⁶ μF, so C=0.0001299*10⁶=130 μF

ehild

 

[chikis]

C = 0.0001299 F is correct. 1 F = 10⁶ μF, so C=0.0001299*10⁶=130 μF

ehild

I really want to understand it, is it
10⁶ or 10^-6

 

[ehild]

"micro" is small. microfarad is a small unit, one millionth, 10^(-6) (1/1000000) of a farad. 1 F = 1000000 microfarad.

ehild

 

[chikis]

"micro" is small. microfarad is a small unit, one millionth, 10^(-6) (1/1000000) of a farad. 1 F = 1000000 microfarad.

ehild

What is the relationship between charge and capacitance. I know of this formula:
Q = CV
where Q = quantity of charge in columb; C = capacitance in uf and V = potential difference between either of the plates in volts.

Suppose C = 2.4uf, can we say that C = 2.4*10^-6 since 1 microfarad uf = 10^-6farad.

By the way one farad is equal to what?
Farad is the unit of what?

 

[StandardBasis]

Why would you say that the voltage in part (i) is 245 volts? I think you should look back at Kirchoff's Rules, unless you can justify that 245 volt answer better.

 

[ehild]

What is the relationship between charge and capacitance. I know of this formula:
Q = CV
where Q = quantity of charge in columb; C = capacitance in uf and V = potential difference between either of the plates in volts.

Suppose C = 2.4uf, can we say that C = 2.4*10^-6 since 1 microfarad uf = 10^-6farad.

By the way one farad is equal to what?
Farad is the unit of what?

Two parallel metal plate can store charge, and when charged, there is a certain potential difference between the plates.

Assume there is +Q charge on one plate and -Q charge on the other plate and the potential difference is V. The ratio C=Q/V is called capacitance.

The unit of capacitance is farad (F). 1 F is the capacitance if 1 coulomb charge is stored at 1 volt potential difference.

C(farad)=Q(coulombs)/V(volts)

1 F = 10⁶μF, or 1μF=10^-6 F. 2.4 μF=2.4 10^-6 F.

ehild

 

[chikis]

Why would you say that the voltage in part (i) is 245 volts? I think you should look back at Kirchoff's Rules, unless you can justify that 245 volt answer better.

The first highlited write up is for a.c circuit:
(1) I got the answer to (i)
as 245V

that of (ii) I got to this extent C= 1/2450(pi)
= 1/7700
= 0.000129870
aprox. 0.0001299
= 1299*10^-4 uf

my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
What really happened?






(2) The second highlited write up is for capacitance of a capacitor:
(2)
What is the relationship between charge and capacitance. I know of this formula:
Q = CV
where Q = quantity of charge in columb; C = capacitance in uf and V = potential difference between either of the plates in volts.

Suppose C = 2.4uf, can we say that C = 2.4*10^-6 since 1 microfarad uf = 10^-6farad.

By the way one farad is equal to what?
Farad is the unit of what?

 

[StandardBasis]

Right, but you should justify why you got the 245 volt across the capacitor...

 

[ehild]

Why would you say that the voltage in part (i) is 245 volts? I think you should look back at Kirchoff's Rules, unless you can justify that 245 volt answer better.

Read the problem and try to find the voltage across the capacitor from the data given. Say if you get different result.

The OP had no problems with question i, he does not need to show his derivation.

ehild

 

[StandardBasis]

Well, I can't get it, but I'd like to solve this problem for my own studying. Ought I start a new thread to do that?

 

[ehild]

Well, I can't get it, but I'd like to solve this problem for my own studying. Ought I start a new thread to do that?

Well, try to solve the problem as you suggested. You have resistor, capacitor and inductor in series. How add up the voltages across them?

ehild

 

[StandardBasis]

I tried thinking of it like a voltage divider-- I found the impedance of the series combination, and tried to take emf*(Zc)/(Ztot), but since we don't know C or L, I couldn't solve it that.

I know that the phase relationship means we can't just straight up add the voltages to 240, but I can't figure out how to get around this.

Although, the same I has to flow through every piece of the circuit the same, so I can find R (14 ohms)... but not L or Q on the capacitor.

Yeah, I can't figure out how to calculate L or C so that I can use the series impedance.

 

[StandardBasis]

All I can think is that I can ignore the other circuit elements, and then I know that Vc=Ic*Zc... Ic=10 Amps, and Zc= 1/(j*w*C). But I don't know C!

 

[ehild]

You can not ignore the other elements.

There are three elements connected in series, and 10 A flowing through one element. What is the current flowing through the others?

The magnitude of the impedance is voltage over current. Z=V/I. What is the resistance of the resistor and what is the impedance of the inductor then?

While the ac voltage across a resistor is in phase with the current, the voltage leads the current by phase pi/2 across an inductor and lags behind the current by pi/2 across a capacitor. You can consider the ac voltages as phasors, and use vector addition to sum them when you apply Kirchhoff's voltage law.

ehild

 

[StandardBasis]

Now I get 90 as the capacitor's voltage. Resistance is 14, inductive reactance is 5, because they each have 10 amps.

I add these impedance as vectors and see that their combined impedance is 14.866. Then 240-(10*14.866)=90 volts left for the capacitor.

Moving in the right direction.

 

[StandardBasis]

Also, I'm realizing that it doesn't make sense for voltage across capacitor to be 245. The max voltage across the source is 240... And that's the max. How could any single element possible have 245 across it, then?

 

[ehild]

Now I get 90 as the capacitor's voltage. Resistance is 14, inductive reactance is 5, because they each have 10 amps.

I add these impedance as vectors and see that their combined impedance is 14.866. Then 240-(10*14.866)=90 volts left for the capacitor.

Moving in the right direction.

Not quite.

There is pi phase difference between the voltages across the capacitor and inductor. So the vector sum of voltages is

V=\sqrt{V_R^2+(V_L-V_C)^2}

You see that the resultant voltage can be smaller than the voltage across either the capacitor or the inductor. Assume, for example, that V_R=50 V and both V_L and V_C are 10000 V, what is the resultant voltage?

The combined impedance of a series RLC circuit is Z=\sqrt{R^2+(Z_L-Z_C)^2}

ehild

 

[StandardBasis]

But I don't know the capacitance, so I can't use Z=-j/wC, and I don't know the voltage obviously, so I can't use Z=V/I.

How about this: I know the source gives V=240sin(wt); the resistor will have in phase current. Thus, its V is ZIsin(wt)=140sin(wt). The inductor's current leads by 90 degrees, so its voltage is ZIcos(wt). The capacitance lags by 90 degrees, so its current is -10cos(wt)/(wC). Is it possible to set those last 3 things equal to the source V?

 

[StandardBasis]

Unless I'm simplifying the leads and lags idea too much. The C and L lead/lag the voltages through THEM by those 90 degrees, but that doesn't mean they lead and lag the voltage source, right? Those could take on different angles than the source as a whole?

 

[ehild]

But I don't know the capacitance, so I can't use Z=-j/wC, and I don't know the voltage obviously, so I can't use Z=V/I.

How about this: I know the source gives V=240sin(wt); the resistor will have in phase current. Thus, its V is ZIsin(wt)=140sin(wt). The inductor's current leads by 90 degrees, so its voltage is ZIcos(wt). The capacitance lags by 90 degrees, so its current is -10cos(wt)/(wC). Is it possible to set those last 3 things equal to the source V?

You do not know the phase of the generator voltage. You can add the voltages across the elements so the generator voltages is

Vg=140 sin(wt)+50cos(wt) -10cos(wt)/(wC) = 240 sin(wt+θ)

You can get wC from this equation.

ehild

 

[StandardBasis]

Why doesn't the resistor have the theta in it as well? Shouldn't it be in phase with the voltage...?

Also, introducing the theta makes it 2 unknowns in one equation, no?

 

[ehild]

The voltage across the resistor is in phase with the current.

The equation holds for any time. What you get when wt=0? What you get if wt=pi/2?
ehild

 

[StandardBasis]

If they are in phase, why doesn't the cosine have the theta, though?

Anyway, plugging in those two times:
50-(10/wC)=240sin(theta)
140=240sin(theta+90)= 240cos(theta)

So I get theta=.948, but that leads to wC being negative...

 

[ehild]

ω can not be negative. cos(theta)=0.5833, and sin theta can be either positive or negative. Choose the negative value.

ehild

 

[SammyS]

Also, I'm realizing that it doesn't make sense for voltage across capacitor to be 245. The max voltage across the source is 240... And that's the max. How could any single element possible have 245 across it, then?

Remember, these are RMS values for voltage.

The sum of the instantaneous voltages is always equal to the instantaneous voltage of the source.

The instantaneous current through any of the devices is the same as through any of the other devices as well as the same through the source.

The instantaneous voltage drop across the resistor is in phase with the instantaneous current.

The instantaneous voltage drop across the capacitor lags behind the instantaneous current by 90°.

The instantaneous voltage drop across the inductor leads the instantaneous current by 90°.

The peak voltage of the source is 240√(2) Volts ≈ 339.4 V .

The peak voltage drop across the capacitor is 245√(2) Volts ≈ 346.5 V

The peak voltage drop across the inductor is 50√(2) Volts ≈ 70.7 V.

At an instant when the instantaneous current is zero, the voltage drop across the resistor is zero. Therefore, the sum of the voltage drops across the resistor, the inductor, and the capacitor is |0 - 346.5 + 70.7 |V ≈ 275.8 V, and this is less than the peak voltage of the source, so there is no conflict there.