• Support PF! Buy your school textbooks, materials and every day products Here!

AC Circuit help

  • Engineering
  • Thread starter James889
  • Start date
  • #1
192
1
Hi,
I have the following circuit:
http://img35.imageshack.us/img35/6735/circuitac.png [Broken]

My task is to calculate the voltage delivered by the voltage source Us
When the voltage over R2 is known, 8cos(100t + 45)

How would you do this?
 
Last edited by a moderator:

Answers and Replies

  • #2
berkeman
Mentor
56,611
6,510
Hi,
I have the following circuit:
http://img35.imageshack.us/img35/6735/circuitac.png [Broken]

My task is to calculate the voltage delivered by the voltage source Us
When the voltage over R2 is known, 8cos(100t + 45)

How would you do this?

Homework Statement





Homework Equations





The Attempt at a Solution

You need to show us your attempts at solving this, before we can be of tutorial help. What have you tried? You can certainly use the KCL and differential equations to solve for that transfer function. Can you think of how to use Phasors to get the solution in an easier way?
 
Last edited by a moderator:
  • #3
192
1
Hi,

Okay i'd try a writing KVL equation for the loop in terms of V.
[tex]
Ri(t) + L(di/dt) -U_{s} = 0[/tex]
[tex]Ri(t) +(1/C)\int i(t)dt +8cos(100t + 45) -U_{s} = 0[/tex]


Does that look reasonable?
 
Last edited:
  • #4
berkeman
Mentor
56,611
6,510
Hi,

Okay i'd try a writing KVL equation for the loop in terms of V.
[tex]
Ri(t) + L(di/dt) -U_{s} = 0[/tex]

<< additional equation deleted by berkeman >>

Does it look reasonable?
Please don't do too much of the OP's work for them. It's fine to provide hints, but generally not good to post equations for the OP here in the Homework Help forums. Thanks.
 
  • #5
192
1
Please don't do too much of the OP's work for them. It's fine to provide hints, but generally not good to post equations for the OP here in the Homework Help forums. Thanks.
Hmm, why did you remove my second equation?
And why did i get a warning for a suggestion to my own problem?
 
  • #6
berkeman
Mentor
56,611
6,510
Hmm, why did you remove my second equation?
And why did i get a warning for a suggestion to my own problem?
Because I'm an idiot. :redface: Sorry. Warning reversed.

Can you re-post your 2nd equation? I don't have a good copy of it in the queue. Thanks!
 
  • #7
berkeman
Mentor
56,611
6,510
Hi,

Okay i'd try a writing KVL equation for the loop in terms of V.
[tex]
Ri(t) + L(di/dt) -U_{s} = 0[/tex]
[tex]Ri(t) +(1/C)\int i(t)dt+8cos(100t + 45) -U_{s} = 0[/tex]


Does that look reasonable?
Oh hey, it's back after reversing the warning. Phew!

And yes, that's a good start.
 
  • #8
192
1
So this is where things start to get tricky.
[tex]R(di(t)/dt) + L(di/dt) = 0[/tex]
[tex]R(di(t)/dt) + RC*(di/dt) + i(t) = 800sin(100t+45)[/tex]

Is that differentiation correct?
 
  • #9
The Electrician
Gold Member
1,246
152
Can you show the intermediate steps instead of just the final result? It seems to me that there should be some second derivatives in there, but I don't see any; check your notation.

Solving simple networks like this with differential equations is really painful; haven't you studied phasors yet?
 
  • #10
192
1
Solving simple networks like this with differential equations is really painful; haven't you studied phasors yet?
we have.
So here's then another suggestion.
Replace the respective element in the circuit with its corresponding complex impedance.

The problem however seems to be that you need the angular frequency of the source(int the formula [tex]-j(1/\omega L)[/tex], which you don't have.
 
  • #11
The Electrician
Gold Member
1,246
152
The angular frequency of the source is the same as the frequency of the voltage across R2, because there aren't any non-linear circuit elements. Does that help you?
 
  • #12
192
1
Yes, thank you
 
  • #13
192
1
The equivalent impedance of the inductance, capacitance and the resistor R2 should be:
[tex]\frac{j\omega L (1/j\omega c + R_{2})}{j\omega L + (1/j \omega c + R_{2})} = 2+2j [/tex]

And now voltage division:
[tex]U_{L} = (2+2j)/(2+2j+R_{1}) * U_{s}[/tex]
[tex]U_{r} = (R_{2}/(R_{2} + (1/j\omgea L)) * U_{s}[/tex]

And we have a Catch 22...
 
  • #14
The Electrician
Gold Member
1,246
152
The equivalent impedance of the inductance, capacitance and the resistor R2 should be:
[tex]\frac{j\omega L (1/j\omega c + R_{2})}{j\omega L + (1/j \omega c + R_{2})} = 2+2j [/tex]

And now voltage division:
[tex]U_{L} = (2+2j)/(2+2j+R_{1}) * U_{s}[/tex]
[tex]U_{r} = (R_{2}/(R_{2} + (1/j\omgea L)) * U_{s}[/tex]

And we have a Catch 22...
This part looks right:
[tex]U_{L} = (2+2j)/(2+2j+R_{1}) * U_{s}[/tex]

But shouldn't the next part be:

[tex]U_{R} = (R_{2}/(R_{2} + 1/j\omega c) * U_{L}[/tex]

When you get done with all this algebra, you should have something of the form:


[tex]U_{R} = TFR * U_{s}[/tex]

where TFR is a transfer function, a complicated algebraic expression.

Then:

[tex]U_{s} = U_{R} / TFR[/tex]
 
  • #15
192
1
Okay, thanks!

One last question though.
Calculating the power delivered to a resistance is pretty simple.
But what is the formula for the power delivered to a current/voltage source?
 
  • #16
The Electrician
Gold Member
1,246
152
For any 2-terminal component, be it a resistor, inductor, capacitor, voltage source, current source, or whatever (or a combination of these between two terminals), the energy absorbed or delivered can be computed by:

[tex]\int e(t)i(t)dt[/tex]

where e(t) is the instantaneous voltage across the terminals, i(t) is the instantaneous current through the terminals, and the limits of the integration are the time period for which you want to know the energy absorbed or delivered. If you divide this energy by the time period involved, then you will get the average power for that time interval.

Typically, for steady state AC problems, the integration is over one cycle. Since for a steady state situation, the average power for any one cycle is the same as for any other cycle, this will be the average power for many cycles, also; it's the steady state average power.
 
  • #17
192
1
For any 2-terminal component, be it a resistor, inductor, capacitor, voltage source, current source, or whatever (or a combination of these between two terminals), the energy absorbed or delivered can be computed by:

[tex]\int e(t)i(t)dt[/tex]
Hi again, about that.

I have the following forumula in my book:
[tex] P = \frac{V_{m} * I_{m}}{2} * \frac{R}{|Z|}[/tex]

I tried to calculate the complex powered delivered to the [tex]V_{1}[/tex] source, but it yield the wrong result.
|Z| = [tex]\sqrt{8^2 + 3.5^2} = 8.73[/tex]

[tex] I_{m} = \frac {V}{|Z|} = \frac{40}{8.73} = 4.58[/tex]
And i already have [tex]V_{m} = 40[/tex]

So consequently i have [tex] \frac{4.58 * 40}{2} * \frac{8}{3.5} = 209[/tex] Which is incorrect.
http://img34.imageshack.us/img34/859/accircuitpng.png [Broken]
 
Last edited by a moderator:
  • #18
The Electrician
Gold Member
1,246
152
To be sure of getting the right answer, rather than using formulas from the book whose applicability is uncertain, I would solve the network. Using the mesh method establish two currents flowing clockwise; I1 in the left mesh and I2 in the right mesh. We then get two equations:

(8 - j*.5)*I1 + (j*.5)*I2 = 40

(j*.5)*I1 + (j*4-j*.5)*I2 = -j*20

Solving, I get:

I1 = 4.9492 + j*.71066 = 5 < 8.17 degrees

I2 = -6.4213 - j*.10152 = 6.422 < -179.09 degrees

So the complex power in V1 is V1 * conjugate(I1) = 197.97 - j*28.43

The complex power in V2 is V2 * conjugate(I2) = 2.0305 + j*128.43

If you add the complex power in V1 and V2 you get 200 + j*100

If you calculate the power dissipated in the 8 ohm resistor you get I12*8 = 200 watts, the same as the real part of the sum of the complex powers from V1 and V2.
 
  • #19
The Electrician
Gold Member
1,246
152
By the way, when I solved the network, I allowed the voltage sources to be sinusoids rather than co-sinusoids, knowing that it wouldn't affect the power calculations. The currents are different, however. To get the currents as you would obtain with cosines of voltage, just multiply the currents I got by "j".
 
  • #20
192
1
Nice, i didn't know you could use the mesh method for calculating power.
 

Related Threads for: AC Circuit help

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
576
Replies
1
Views
2K
Replies
2
Views
931
Replies
1
Views
963
  • Last Post
Replies
3
Views
830
  • Last Post
Replies
7
Views
2K
Top