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

  • Thread starter Thread starter blue_tiger30
  • Start date Start date
  • Tags Tags
    Load Ohm Voltage
Click For Summary

Discussion Overview

The discussion revolves around finding the open-circuit voltage (Voc) and the voltage across a 5 ohm load (VL) using three different methods: a direct approach with potential dividers, Thévenin's theorem, and nodal analysis. Participants are sharing their attempts and seeking clarification on their calculations and methods.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant reports using the direct approach and obtaining Voc: 98.63 e^-j3.38 and VL: 98.45 e^-j6.810, but struggles to replicate these results using the other methods.
  • Another participant questions the correctness of the results obtained from the direct method and asks for more detailed calculations to verify them.
  • One participant describes their calculations for the Thévenin equivalent, arriving at a voltage source of 99.15 volts and a series resistance of j0.597Ω, leading to a load voltage of 97.76 -j11.67 volts.
  • Participants express uncertainty about the nodal analysis method, with one noting that their results differ from those obtained through the direct approach and Thévenin's theorem.
  • There is a suggestion that the answers from the different methods should be equivalent, but participants are unsure where their calculations may have gone wrong.
  • One participant acknowledges a lack of understanding in nodal analysis and requests corrections to their approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of their calculations, particularly regarding the nodal analysis method. There is acknowledgment that the results from the direct approach and Thévenin's theorem should align, but discrepancies remain in the calculations.

Contextual Notes

Participants express uncertainty about their mathematical steps, particularly in the nodal analysis, and there are indications of missing assumptions or potential errors in the calculations. The discussion reflects varying levels of understanding of the methods involved.

blue_tiger30
Messages
28
Reaction score
0

Homework Statement


https://dl.dropbox.com/u/3405118/24-02-2013%2018-46-54.png

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

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:
Physics news on Phys.org
blue_tiger30 said:

Homework Statement


https://dl.dropbox.com/u/3405118/24-02-2013%2018-46-54.png

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


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
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:
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
 
blue_tiger30 said:
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
Okay, looks good.
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
Yes, your Thevenin components look fine, and your result for the load voltage looks good.
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
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?
 
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 don't really get nodal analysis I only know how to do loop analysis but that's not what he asked for in this homework so If you could correct my steps if possible

many thanks for your help
 
blue_tiger30 said:
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 ?
The answers are the same; one is expressed in polar form, the other in rectangular form :wink:
to be honest I don't really get nodal analysis I only know how to do loop analysis but that's not what he asked for in this homework so If you could correct my steps if possible

Your node equations look fine, but in the simultaneous solving of them something went wrong. So check your math in the last two lines.
 

Similar threads

Engineering Why is the voltage source high potential into node 1 at -12 V in this op amp problem?
  • · Replies 15 ·
Replies
15
Views
3K
Engineering Calculating the voltage at a certain Resistor
  • · Replies 3 ·
Replies
3
Views
2K
Engineering Calculating Norton Equivalent Circuit for Complex Load Impedance
  • · Replies 2 ·
Replies
2
Views
2K
Engineering Source impedance using given voltage swing and power
  • · Replies 2 ·
Replies
2
Views
2K
What is current across 5 ohm resistor
  • · Replies 3 ·
Replies
3
Views
8K
Engineering Norton/Thevenin Equivalent circuits with a dependent voltage source
  • · Replies 9 ·
Replies
9
Views
6K
Finding Steady State Amplitude of Voltage Across Load Resistor
  • · Replies 3 ·
Replies
3
Views
3K
Engineering Solve RL Parallel Circuit: Find VL(0) Across 5ohm Resistor
  • · Replies 13 ·
Replies
13
Views
3K
Engineering Finding the Thevenin Equivalent Circuit for a Network with a Load Resistor
  • · Replies 1 ·
Replies
1
Views
2K
Ohm's Law - Finding Source Voltage
  • · Replies 3 ·
Replies
3
Views
2K