Thevenins theorem Query

Discussion in 'Homework Help' started by Kev 101, Apr 7, 2014.

1. Kev 101 Thread Starter New Member

Apr 7, 2014
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Recently was given a circuit to analyse, the method we had to use was thevenins to get the thevenins resistance and the thevenins voltage.

Then the question asks to find the voltage drops, current and power dissipated through the other resistors. three resistors in total.

My question is which process do we use to work this out?

would it be simpler if i uploaded the image with the questions so it is seen clearly.

Thank You

Best Regards

Kev

2. Kev 101 Thread Starter New Member

Apr 7, 2014
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just too add to the previous post which i had forgotten to mention is that i have already worked out the thevenin values and have re-drawn the equivalent circuit it is just the second part that is causing me confusion.

3. Kev 101 Thread Starter New Member

Apr 7, 2014
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The images are of the work i have currently done regards to my first post.

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4. WBahn Moderator

Mar 31, 2012
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Your Thevenin Equivalent circuit is correct. Normally the equivalent resistance is placed in the top leg, but it really doesn't matter.

Using the equivalent circuit, you can ONLY find the voltage and current for the load. You can't found out ANYTHING about the original components because the equivalent circuit is ONLY equivalent in terms of the voltage and current relationships at the terminals.

But once you have the voltage and the current for the load, do you have enough information to find the current in R2? With that current and the load voltage, can you find the voltage at the junction of the three resistors relative to the bottom node? With that can you find the current in R3?

See how you just walk things along until you have all the values worked out?

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5. shteii01 AAC Fanatic!

Feb 19, 2010
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I would use either Node Voltage Method or Mesh Current Method.

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6. Kev 101 Thread Starter New Member

Apr 7, 2014
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Thank you for the help so far guys it is much appreciated.

I have to admit that this subject of my course i have only just begun, and the tutor that we have teaching the subject, well lets just say all students would benefit with a bit more lecture time to carry on with the assignments she has handed out.

not to moan and complain as i would rather just get on with it.

we did get shown mesh analysis briefly, well i say briefly it was more of a work sheet which was not printed very well that we had to work through. how does mesh analysis work with just the one voltage source in a single loop (as in the above circuit).

P.S i am not looking for you guys to give me the answer as this is my work. Just for a better understanding in order to complete the task which is set.

Thank You.

7. WBahn Moderator

Mar 31, 2012
18,061
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The key to remember is that Mesh Current Analysis is nothing more than a formalized and systematic way of applying KVL around a set of easily identified loops. Similarly, Node Voltage Analysis is nothing more that a formalized and systematic way of applying KCL at a set of easily identifiably nodes (junctions).

If you have just one loop with a single voltage source, then mesh analysis reduces to essentially Ohm's law.

If you have a voltage source V1 in series (i.e., a single loop) with R1 and R2 (use your diagram to set the polarity on V1), then you declare which direction your loop current goes (I almost always use a loop current that flows counter-clockwise) called I1. Now go around the loop and sum up all of the voltage drops in the direction of the loop current. You get I*R1 + I*R2. That goes on the left hand side of the equation. Now go around the loop again and sum up all the voltage gains in the direction of the loop. You get -V1. That goes on the right side of the equation. So your loop equation is

I1*R1 + I1*R2 = -V1

Solve for I1

I1 = -V1/(R1+R2)

So the actual current is flowing clockwise (because of the minus sign) and the magnitude is the voltage divided by the sum of the resistances, which is simply Ohm's Law.

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8. Kev 101 Thread Starter New Member

Apr 7, 2014
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I1 = -V1/(R3+R4)
I1 = -48/(12000+4000)
I1 = 3mΩ

is this correct.

voltage drop at R4 would then be:

VR4 = I1*R4
VR4 = 3mΩ*12kΩ
VR4 = 36 V

Last edited: Apr 8, 2014
9. WBahn Moderator

Mar 31, 2012
18,061
4,904
Check the units!

Does a current of 3mΩ make any sense?

Does a mΩ*kΩ yield volts?

What happened to the minus sign?

10. Kev 101 Thread Starter New Member

Apr 7, 2014
19
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rather simple mistake on my part the calculation should have been amps for the 3m (-3mA) not ohms.

VR4 = I1*R4
VR4 = -3mA*12kΩ
VR4 = -36 V

11. WBahn Moderator

Mar 31, 2012
18,061
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Better.

Now, whether VR4 is -36V or +36V depends on the assigned polarity of the voltage across VR4, which no one has defined. So you need to mark on the diagram which end of R4 is the positive end and then, depending on that assignment, you will have either:

VR4 = I1*R4

or

VR4 = -I1*R4

It will be the first if I1 (the symbolic I1, not the actual current) enters the positive terminal (which would be the case if the right side of R4 is assigned as the positive terminal) and the second if I1 leaves the positive terminal (which would be the case if the left side of R4 is assigned as the positive terminal).

12. Kev 101 Thread Starter New Member

Apr 7, 2014
19
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I am very thankful for you guiding me through this.

the positive side of R4 will be the left hand side.

what will be the next steps in order for me to complete the task.

Thank You

13. shteii01 AAC Fanatic!

Feb 19, 2010
3,496
511
* find voltages across all the resistors
* find currents through all the resistors
* find power dissipated by each resistor

So.
* you have 3 currents
* you have 4 voltages
* for power you can use:
P=V(I) or P=R(I^2) or P=(V^2)/R your choice

So far you found I1 and VR4, I1=3 mA, VR4=36 V.

You need 2 more currents, 3 more voltages and 4 power calculations.

14. Kev 101 Thread Starter New Member

Apr 7, 2014
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I need to find the currents for I2 and I3.
voltage drops for R3 and R2
Power dissipated in R4, R3 and R2

Thank You

15. WBahn Moderator

Mar 31, 2012
18,061
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The first thing to do is annotate the diagram with the assigned voltage polarities for each device. Just pick one side of each resistor and put a + next to it. Then, whenever you talk about VR3 we know that it is the voltage across VR3 such that if VR3 > 0 then the side with the + is more positive than the other side. Similarly, If IR3 > 0 then the current is flowing into the side with the +.

Next, pick a reference node. You get to call one node (your choice) your common node and assign it a voltage of 0V. By convention, the node connected to the negative side of the source is the most obvious choice.

Now you can walk from node to node computing the voltage at each node (relative to the common node). Any time you know the voltage at both nodes connected to a resistor, you can calculate the current in that resistor. Any time you know the voltage at one node of a resistor as well as the current through that resistor, you can calculate the voltage at the node at the other end of the resistor.

16. WBahn Moderator

Mar 31, 2012
18,061
4,904
Oh, and remember that the circuit we just were looking at was NOT the circuit in the problem. The circuit in the problem has two loops due to the presence of the load resistor.

The circuit we were looking at was in response to you asking about how you would deal with mesh analysis for a circuit with a single loop and one source. The only way that this applies to this circuit is to find Vth, which you've already done successfully.

17. shteii01 AAC Fanatic!

Feb 19, 2010
3,496
511
Either Node Voltage Method or Mesh Current Method. Pick one.

18. Kev 101 Thread Starter New Member

Apr 7, 2014
19
0
shteii01
i have not done much with mesh or node analysis, the only task we had was on mesh analysis with two voltage sources, i would not know where to begin with the cicuit in the task.

Thank You

19. Kev 101 Thread Starter New Member

Apr 7, 2014
19
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This was the mesh task which was set out with a different circuit

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20. Kev 101 Thread Starter New Member

Apr 7, 2014
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where as this is the circuit that has been set in the task.

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