thevenin and norton rlc circuits

Thread Starter

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
 

Thread Starter

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
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.
 

Thread Starter

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

?!?!?
 

Thread Starter

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?
 
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