# Solve Superposition Theorem for R1, R2 Circuit

• Engineering
• leejohnson222
In summary, The conversation discusses the process of finding the current and voltage at various resistors in a circuit by using superposition. The steps involved include using Norton/Thevenin source transformations, assigning variables to each source value, and calculating the currents through each resistor. The direction of the current is important, and the final step is to sum the currents while taking into account the directions. The conversation also mentions other methods such as Millman's theorem, but focuses on the use of superposition.
leejohnson222
Homework Statement
find current and voltage at r1, r2
Relevant Equations
KCL
i have this question and i just want to make sure i am on the right track as i know there are quite a few steps to this to get to the final soultions

i need to find Current at r1
voltage at r1
current at r2
voltage at r2

i have so far split this into 2 drawings so i am dealing with one power source at a time,
taking r1 and r2 as parallel i found 1/RT to be 2 ohms
so current being 30/2 = 15amps
voltage at r1 current x resistance = 15 x 3 = 45v
i understant now to get the complete answer i would need to do the same with the other power source and combine the figures
Is this correct or have i missed a step out ?

Revisit how you arrive at R1 and R2 becoming parallel, when zeroing out one of the sources. Are you able to show your other drawings?

Yes, please show us what you mean with some schematics.

Superposition doesn't work if you change the passive parts of the network. i.e. you can't add, move, remove, or change the value of the resistors. Where it does work is with setting all sources (voltage or current sources) to zero and solving the same network, with all sources still present but mostly having zero value. Then if you solve for each source value independently, you can add up the separate results to get the correct solution when all of the sources are active.

scottdave
yes this what i was going to do, work out both sides of this with different power sources
i havent drawn any thing i was going to make of results for each side

leejohnson222 said:
yes this what i was going to do, work out both sides of this with different power sources
i havent drawn any thing i was going to make of results for each side
You should be able to have one schematic, since the network doesn't change. Just assign variables to each source value. Then sometimes those variables are zero in the analysis, sometimes they are non-zero.

scottdave
Note that you need to calculate the currents through resistor 3 to solve this, even though the question does not ask for it. It appeared that you were ignoring R3 in your initial post.

Also remember that the direction of each current is important. For example (I made up these numbers) if R1 has 10 amps "up" in one calculation, then 12 amps "down" in the other calculation, the superposition of these is 2 amps down.

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DaveE
Millman theorem it is easier, I think

Millman theorem it is easier, I think
Yes. But the question was about Superposition, I think.

I would have used Norton/Thevenin source transformations myself. Millman's theorem never excited me. There are several approaches to choose from.

I wouldnt ignore R3 as when you short circuit the E2 R3 would be in series while R1, R2 are in
but when you short circuit E1, im not 100% sure i think 2 loops are created one going left and one going right, from positive to negative returning to the power source

to avoid making it too confusing i could draw 2 scematics to show each power source being used with values, then combine the two results, that is what i have read about the way to do this so far

scottdave
You have to put E1=0 and short the source and then calculate the currents as if you only had one source. Do the same with E2. And finally, sum the currents on each branch.

scottdave
You have to put E1=0 and short the source and then calculate the currents as if you only had one source. Do the same with E2. And finally, sum the currents on each branch.
And when you "sum" the currents, the directions of the currents are important.

yes i understand that directions need to be taken into account at the end so you get the correct voltage depending on direction of each power source.

so far i have this for 45v source
r1//r1 = 2ohms
r3 = 3 oms
Current Total is 45v/5 = 9amps
then i need to find each current through each resistor

leejohnson222 said:
yes i understand that directions need to be taken into account at the end so you get the correct voltage depending on direction of each power source.

so far i have this for 45v source
r1//r1 = 2ohms
r3 = 3 oms
Current Total is 45v/5 = 9amps
then i need to find each current through each resistor
Yes (except for the typo r1||r1...). Next you want to solve for the node voltages for this case.

SammyS
great sorry about typo, one question, when i am lookin at the 2nd voltage are all resistors all in series? since voltage will flow in 2 directions would any resistors be in parallel?
if i label the top node as node a voltage i have is 18v ? not sure i have this correct

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Voltage does not "flow".
When E1 is shorted (0 volts) and only E2 is present, you only have to verify how (which way) the current - driven by E2 - will go. Through all the resistors - one after another? Or does it take different ways in parallel?

sorry i mean current flow, when E2 is active and E1 is shorted out i looks to me like current will flow left and right, creating 2 loops, but if that is true then would R1 and R3 always be in parallel with R2 on each loop?

When E1 is shorted, you can say that R1 is parallel to R3, since it shares a node at both top and bottom. When E1 is active, then R1 only shares 1 node with R2 or R3 so it is not parallel.

I hope that explanation makes sense.

Similar when talking about R2, when you short E2.

leejohnson222 said:
if i label the top node as node a voltage i have is 18v ? not sure i have this correct

Yes. You have 9A flowing through R1, a 3Ω resistor, which will make a voltage drop of 9A⋅3Ω = 27V. You can subtract this from the E1 voltage drop, 45V (Note the polarities are really important!) to get a total drop across the E1-R1 branch of 18V.

The alternate way is to see that the 9A also flows through R2||R3, which is 2Ω. So that will make an 18V drop across those two resistors.

This is all with E2 set to 0V, of course.

scottdave
thank you so much, this makes quite a bit of sense now and i was able to get closer to my answers and follow why i was doin it also, thanks again

DaveE

## What is the Superposition Theorem?

The Superposition Theorem states that in a linear electrical circuit with multiple independent sources, the voltage across or current through any element in the circuit is the algebraic sum of the voltages or currents produced by each independent source acting alone, with all other independent sources replaced by their internal impedances (usually a short circuit for ideal voltage sources and an open circuit for ideal current sources).

## How do you apply the Superposition Theorem to a circuit with resistors R1 and R2?

To apply the Superposition Theorem to a circuit with resistors R1 and R2, follow these steps: 1. Consider one independent source at a time while replacing all other independent sources with their internal impedances.2. Solve the circuit to find the contribution of the active source to the voltage or current of interest.3. Repeat the process for each independent source.4. Sum the contributions from all independent sources to find the total voltage or current in the circuit.

## What happens to the other sources when considering one source at a time in the Superposition Theorem?

When considering one source at a time in the Superposition Theorem, all other independent voltage sources are replaced by short circuits (effectively a wire with zero resistance), and all other independent current sources are replaced by open circuits (effectively a break in the circuit with infinite resistance).

## Can the Superposition Theorem be applied to non-linear circuits?

No, the Superposition Theorem cannot be applied to non-linear circuits. It is only valid for linear circuits, where the principle of superposition holds true. This means the circuit components must obey Ohm's Law, and the responses (voltage and current) must be directly proportional to the sources.

## What are the limitations of the Superposition Theorem?

The Superposition Theorem has several limitations: 1. It is only applicable to linear circuits.2. It does not work for power calculations directly, as power is a non-linear function of voltage and current.3. It can be time-consuming for circuits with many sources, as each source must be considered individually.4. Dependent sources are not turned off and must be left as they are while applying the theorem.

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