# thevenin and norton rlc circuits

Discussion in 'Homework Help' started by rents89, Feb 5, 2009.

1. ### rents89 Thread Starter Member

Feb 5, 2009
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0
any idea on how to solve these??

cheers

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Did you read the sticky post? You must show you're own work as far as you've been able to get before anyone will help you.

3. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
yeh sorry didnt read it, i have done some calculations, but i am unsure about the method.. was wonderin if anyone could point me in the right direction...? thanks alot

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Tell us the method you've been trying to use.

Feb 4, 2008
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6. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
ok
for q1:

- tried to find thevenin's impedance (Z_Th)
-then calculated V_Th
-finally V_Th/(Z_Th - j40) to find the current in the capacitor.

q2:

-converted voltage source to current and got 0.375<0° A
-rearranged the circuit and trying to find Z_Norton

pretty stuck basically...
cheers guys

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Start with the very first part. If you'll show your calculations for finding Zth, then we can show you where you went wrong.

8. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
Z_th= ((50+j50)(30))/((50+j50)+(30))

but im pretty sure its wrong

9. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
do i need to add anything else? cheers

10. ### mik3 Senior Member

Feb 4, 2008
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No, you are correct.

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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So far, so good. Now show the calculation for Vth. Show the final result.

Then show the final number you get for the current in the cap as a single complex number, but also show the steps to get that result.

12. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
really?? ok, so how to calculate V_Th?
I got,

((50+j50)/(80+j70))*200<0°

13. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
sorry just saw the post,

V_Th= 133.04< -141.70°

I= ((V_Th/(Z_Th - j40))= 3.31<-88.49°

=2.86+1.66i

?!?!?

14. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
helppppp meeee

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I think this should be Vth = ((50+j50)/(80+j50))*200<0°

16. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
safe cheers, is the method for calculating the current in the cap correct?

also any help on q2? cos i converted the voltage source to current, is that right?

thanks alot man

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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You have Vth wrong; I showed the mistake in the previous post.

Vth = 149.91<12.995°

Then I = ((Vth/(Zth-j40)) = ((149.91<12.995°/(22.486<12.995°-j40) =
3.6346<70.907° = 1.1889 + j3.4346

I think your only mistake was getting Vth wrong.

Have you attempted the second problem yet?

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Once again, it's difficult to help if you don't show all your steps; that's the only way I can see if you made a mistake somewhere, and just where you made it.

19. ### rents89 Thread Starter Member

Feb 5, 2009
14
0
yeh sorry once again, my attempt at converting the voltage source is on the diagram... am i on the right track?

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Feb 4, 2008
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Yes you are.