Superposition Theorem help please

2

Hi,
I have spent a lot of time this weekend trying to get my head around Superposition Theorem, I am hoping I am there with it, but could anyone kindly have a look through and tell me if the workings look correct, I am ending up with a negative current above R3 and not sure how to flip it in my calcs.
For E1 at R3 I am getting 1A, and E2 at R3 I am getting -2.25A, which is giving me -1.25A.
I thought that when you separate the sources you keep the current flow the same for the first one and second one, even though E2 wouldn't go clockwise.

First time posting, so not sure what the etiquette is about uploading answers.

Thanks

Reply 1

Nitrotoluene

12

Original post

by slimtimuk

Hi,
I have spent a lot of time this weekend trying to get my head around Superposition Theorem, I am hoping I am there with it, but could anyone kindly have a look through and tell me if the workings look correct, I am ending up with a negative current above R3 and not sure how to flip it in my calcs.
For E1 at R3 I am getting 1A, and E2 at R3 I am getting -2.25A, which is giving me -1.25A.
I thought that when you separate the sources you keep the current flow the same for the first one and second one, even though E2 wouldn't go clockwise.
First time posting, so not sure what the etiquette is about uploading answers.
Thanks

Hello slimtimuk!
Alright, this is a superposition problem with two DC power supplies in the circuit.
Here's how to solve it:
-First, turn on only power supply E1, and short E2 (make it a wire).
-Second, do the opposite. Turn on only power supply E2, and short E1 (make it a wire).
-Third, add the results to get the final answer .
For each part, work out the current and voltage for each resistor.
Then, put all of the numbers together to get the total current and voltage for each resistor.
Here is a resume of the math:
Part 1: E1 only (E2 shorted)
R1 and R2 are in series, so the total resistance is 6Ω.
The current from E1 (I1') is 8V / 6Ω = 1.33A.
The voltage across R1 (V1') is 2.67V.
The voltage across R2 (V2') is 5.33V.
R3 has no current, so I3' = 0A and V3' = 0V.
Part 2: E2 only (E1 shorted)
Again, R2 and R3 are in series, so the total resistance is 8Ω.
The current from E2 (I2'') is 12V / 8Ω = 1.5A.
The voltage across R3 (V3'') is 6V.
The voltage across R2 (V2'') is 6V.
R1 has no current, so I1'' = 0A and V1'' = 0V.
Part 3: Use the results above. Final result:

1.

Current through R1 (I1): 1.33A + 0A = 1.33A.

2.

Voltage across R1 (V1): 2.67V + 0V = 2.67V.

3.

Current through R3 (I3): 0A + 1.5A = 1.5A.

4.

Voltage across R3 (V3): 0V + 6V = 6V.

5.

Voltage across R2 (V2): 5.33V + 6V = 11.33V.

6.

Power used by R2 (P2): (11.33V)^2 / 4Ω = 32.11W.

See ya!
Sandro

Reply 2

Eimmanuel

Study Forum Helper

15

Original post

by slimtimuk

Hi,
I have spent a lot of time this weekend trying to get my head around Superposition Theorem, I am hoping I am there with it, but could anyone kindly have a look through and tell me if the workings look correct, I am ending up with a negative current above R3 and not sure how to flip it in my calcs.
For E1 at R3 I am getting 1A, and E2 at R3 I am getting -2.25A, which is giving me -1.25A.
I thought that when you separate the sources you keep the current flow the same for the first one and second one, even though E2 wouldn't go clockwise.

First time posting, so not sure what the etiquette is about uploading answers.

Thanks

You can read the posting guidelines here or follow the link below:
https://www.thestudentroom.co.uk/showthread.php?t=6736138

I agree with your answer for the current in R3.
In using superposition theorem, we short E2, to find the current in R3 due to E1.
You are right that the current is 1.00 A and the direction is from top to bottom.
Then we short E1, to find the current in R3 due to E2.
You are right that the current is 2.25 A and the direction is from bottom to top.

The superposed current in R3 is
2.25 - 1.00 = 1.25 A (flows from bottom to top)

Can always check your answer using this simulation.
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc-virtual-lab/latest/circuit-construction-kit-dc-virtual-lab_all.html

Spoiler

Reply 3

slimtimuk

OP

2

Original post

by Eimmanuel

You can read the posting guidelines here or follow the link below:
https://www.thestudentroom.co.uk/showthread.php?t=6736138
I agree with your answer for the current in R3.
In using superposition theorem, we short E2, to find the current in R3 due to E1.
You are right that the current is 1.00 A and the direction is from top to bottom.
Then we short E1, to find the current in R3 due to E2.
You are right that the current is 2.25 A and the direction is from bottom to top.
The superposed current in R3 is
2.25 - 1.00 = 1.25 A (flows from bottom to top)
Can always check your answer using this simulation.
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc-virtual-lab/latest/circuit-construction-kit-dc-virtual-lab_all.html

Spoiler

Thanks very much for the help, I did try it on a simulator and got 1.25A.
I simulated both E1 and E2 separately as well.
My incorrect calculations are because I have ended up with E1 - E2.
Am I correct in that E1 is flowing in a different direction to E2, that means I have this ? 1 - 2.25=-1.25.
Not sure how I can change this around apart from doing as you have with E2-E1 which is the correct answer.

Reply 4

Eimmanuel

Study Forum Helper

15

Original post

by slimtimuk

Thanks very much for the help, I did try it on a simulator and got 1.25A.
I simulated both E1 and E2 separately as well.
My incorrect calculations are because I have ended up with E1 - E2.
Am I correct in that E1 is flowing in a different direction to E2, that means I have this ? 1 - 2.25=-1.25.
Not sure how I can change this around apart from doing as you have with E2-E1 which is the correct answer.

I would not say that “E1 - E2 = 1 - 2.25 = -1.25” is incorrect calculation. You need to understand the physics. You are assuming E1 to give the current flow in the positive direction while E2 produces a negative direction of current.
The minus sign in the answer is indicating your choice of positive current flow direction is opposite of your answer.
You may need to consult your notes or textbook on how to solve circuit problems systematically and interpret your answer properly.

Superposition Theorem help please

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