# thevenin and norton rlc circuits

#### rents89

Joined Feb 5, 2009
14
any idea on how to solve these??

cheers

#### The Electrician

Joined Oct 9, 2007
2,778
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.

#### rents89

Joined Feb 5, 2009
14
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

#### The Electrician

Joined Oct 9, 2007
2,778
Tell us the method you've been trying to use.

#### rents89

Joined Feb 5, 2009
14
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

#### The Electrician

Joined Oct 9, 2007
2,778
Start with the very first part. If you'll show your calculations for finding Zth, then we can show you where you went wrong.

#### rents89

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

but im pretty sure its wrong

#### rents89

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

#### mik3

Joined Feb 4, 2008
4,843

#### The Electrician

Joined Oct 9, 2007
2,778
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.

#### rents89

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

((50+j50)/(80+j70))*200<0°

#### rents89

Joined Feb 5, 2009
14
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.
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

?!?!?

#### rents89

Joined Feb 5, 2009
14

#### The Electrician

Joined Oct 9, 2007
2,778
really?? ok, so how to calculate V_Th?
I got,

((50+j50)/(80+j70))*200<0°
I think this should be Vth = ((50+j50)/(80+j50))*200<0°

#### rents89

Joined Feb 5, 2009
14
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

#### The Electrician

Joined Oct 9, 2007
2,778
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

?!?!?
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?

#### The Electrician

Joined Oct 9, 2007
2,778
also any help on q2? cos i converted the voltage source to current, is that right?
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.

#### rents89

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

#### mik3

Joined Feb 4, 2008
4,843
Yes you are. 