# Find the oc voltage and the voltage across the 5 ohm load

1. Feb 24, 2013

### blue_tiger30

1. The problem statement, all variables and given/known data
https://dl.dropbox.com/u/3405118/24-02-2013%2018-46-54.png [Broken]

Find (a) the open-circuit voltage VL , and (b) the voltage across the 5 ohm load, by the following methods.

Method 1: Direct approach. First find the open circuit voltage by simple ‘potential divider’ approach
(but with complex impedances Z replacing the simple resistances we used under d.c. conditions). Then
find the equivalent series impedance of the two parallel branches on the right, and again use the ‘potential
divider’ idea to find VL. The calculations are lengthy, and you will probably find it best to use polar form
for the multiplications and divisions.

Method 2: Thévenin. Find the Thévenin equivalent circuit for the transformer and 100 V supply. Then
apply the load and use the potential divider approach.
Method 3: Nodal Analysis. Call the node at the top of j35 node a and that at the top of the 5  load
node b, with the reference (zero) node at the bottom. Write down the nodal equations in terms of the
complex impedances, and solve for Vb.

no drawings required

3. The attempt at a solution

I did method one and got the answer to be : Voc :98.63 e^-j3.38 , VL: 98.45 e^-j6.810

and tried to do the rest but I couldn't get the same answer can some one explain how do we the same answer for method 2 and 3

Last edited by a moderator: May 6, 2017
2. Feb 24, 2013

### Staff: Mentor

Hi blue_tiger30, Welcome to Physics Forums.

How do you know that the result you obtained by method 1 is correct? Did you get different values for both methods 2 and 3?

Can you show more detail of your work for method 1 so that we can check it?

Last edited by a moderator: May 6, 2017
3. Feb 24, 2013

### blue_tiger30

in the start I replaced the resistances that I got into a complex impedance z = ((5+j0.3)*j35))/(j35+4(5+j0.3)=4.818+j0.9804ohm
I=v/(z+j0.3)
Voc=z*I=v*(z/(z+j0.3))= 98.63 e^-3.38
V across R = I*5=(Voc/(5+j0.3))*5= 98.45 e^-j6.810
I think this is th right answer becuse I UNDERSTAND THIS METHOD
but for the rest I did as follows :
The Thevenin equivalent is
a voltage source = 100(j35/(j35 +j0.3)= 100(j35/j35.3) = 99.15 volts, this is the open circuit voltage
in series with R thevenin = j0.3 + (j0.3)(j35)/(j0.3 + j35) = j0.597Ω

The voltage across RL is 99.15( 5/(5 + j0.597)) = 97.76 -j11.67 volts

Nodal analysis
Node A: (100 - Va)/(j0.3) = Va/(j35) + (Va - Vb)/(j0.3)
100/j0.3 = 2Va/j0.3 + Va/j35 - Vb/j0.3
100 = 2Va + Va(0.3/35) - Vb = Va( 2.0086) -Vb

Node B: (Va - Vb)/(j0.3) = Vb/5
Va/j0.3 = Vb/j0.3 + Vb/5
Va = Vb + Vb(j0.3/5) = Vb( 1 +j0.06)

100 = (Vb(1 +j0.06)(2.0086) -Vb = Vb(1.0086 + j0.121)
Vb = 100/(1.0086 +j0.121) = (100.86 - j12.1)/1.016 = 99.3 -j11.9 volts

which give me different answers and sice I m not that good at these methods I thinks that I did them in the wrong way

4. Feb 24, 2013

### Staff: Mentor

Okay, looks good.
Your "Direct Approach" and Thevenin methods both look fine, but things have gone wrong with your Nodal Analysis.

When doing Nodal Analysis you should first choose a reference node which is defined to have zero potential, then identify the independent nodes. It doesn't look like you've done this. The circuit has only two nodes of interest: a and b. The rest of the junctions lie along series connected branches. Which one will you take to be the reference node?

5. Feb 24, 2013

### blue_tiger30

so for method 2 shouldn't the answer be = to the answer I got in method one ? can u please tell me what went wrong there ?

to be honest I dont really get nodal analysis I only know how to do loop analysis but thats not what he asked for in this hw so If you could correct my steps if possible

6. Feb 24, 2013

### Staff: Mentor

The answers are the same; one is expressed in polar form, the other in rectangular form
Your node equations look fine, but in the simultaneous solving of them something went wrong. So check your math in the last two lines.