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

In summary: Yes, the answer for method 2 should be the same as the answer for method 1. It looks like you have made a mistake in your calculation for the Thevenin equivalent components. The open circuit voltage should be 99.15 volts, not 99.3 volts. This small difference in voltage can lead to a significant difference in the final result.As for the nodal analysis, it looks like you have made a mistake in setting up the equations. Here is one possible way to do it:Node a: (100 - Va)/(j0.3) = Va/(j35) + (Va - Vb)/(j0.3)Node b: (Va - Vb)/(j0.
  • #1
blue_tiger30
29
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:
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  • #2
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:
  • #3
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
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?
 
  • #5
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
 
  • #6
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.
 

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

1. What is an oc voltage?

The oc voltage, also known as the open circuit voltage, is the voltage measured across a circuit when there is no load or current flowing through it. It is the maximum voltage that can be obtained from a power source.

2. How do you find the oc voltage?

To find the oc voltage, you need to disconnect the load from the circuit and measure the voltage across the open ends using a voltmeter. This will give you the oc voltage of the circuit.

3. What is the significance of finding the oc voltage?

The oc voltage is important in determining the maximum voltage that a circuit can produce. It is also useful in analyzing the characteristics and limitations of a power source.

4. What is the voltage across a 5 ohm load?

The voltage across a 5 ohm load can be calculated using Ohm's Law, which states that voltage (V) is equal to the product of current (I) and resistance (R). So, if the current through the load is known, the voltage can be calculated using V = I x R.

5. How does the oc voltage affect the voltage across a 5 ohm load?

The oc voltage does not directly affect the voltage across a 5 ohm load. However, the oc voltage is used to determine the maximum voltage that a circuit can produce, which in turn can affect the voltage across the load if the circuit is operating at its maximum capacity.

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