# quick/easy superposition question

#### ihaveaquestion

Joined May 1, 2009
314
hi people,

quick question... what is the voltage at the node marked V (using superposition)

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#### ihaveaquestion

Joined May 1, 2009
314
this is the teacher's solution:

shorting the right hand side voltage source, you have the 1 and 3 resistors in parallel and you have a voltage divider to find the voltage from the left side source.

shorting the left hand side voltage source, the teacher again used voltage division and said its 3 parallel with 2, which i agree with... so the voltage due to that source is 5*3||2/( 5 + 3||2)... this i dont agree with

i have a attached another picture showing that if we wanted the voltage where i have marked "NOT HERE" then i would say voltage divison makes sense... but if we want the voltage where V is, why isnt it just 5 volts due to the right hand source with left hand shorted.

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#### t_n_k

Joined Mar 6, 2009
5,455
shorting the left hand side voltage source, the teacher again used voltage division and said its 3 parallel with 2, which i agree with... so the voltage due to that source is 5*3||2/( 5 + 3||2)... this i dont agree with
You're right ...

It should be 5*3||2/(1+3||2)

#### ihaveaquestion

Joined May 1, 2009
314
I meant to type that tnk sorry.. that's what the teacher put... could you give me an explanation of why that is it... that isn't exactly the voltage divison relationship so i am quite confused...

thanks

#### t_n_k

Joined Mar 6, 2009
5,455
I meant to type that tnk sorry.. that's what the teacher put... could you give me an explanation of why that is it... that isn't exactly the voltage divison relationship so i am quite confused...

thanks
The circuit should have a reference (or ground) point indicated - let's say that is the bottom node of the diagram. VA is then given with respect to the reference node. You've denoted the reference as 'Here'.

Looking at the second diagram the re-arrangement drawn is correct. A 5V source is connected in series with 3||2Ω and a 1Ω. The source voltage is divided across the series combination with the parallel branch {2||3Ω} voltage being given according to the previously posted equation.

i.e.

V 2||3Ω combination = 5*3||2/(1+3||2)

The superposition value for VA in this case is the value actually given above - since the 2||3Ω parallel combination sits astride Node A and the circuit reference point.

#### ihaveaquestion

Joined May 1, 2009
314
I suppose I need help with the basics then...

In the attached picture, I have a few questions:

I would think that the voltage at V1 is simply V (the source) and at V2 minus V (-V), but apparently this isn't the case.

What are V1 and V2? What's the difference between V1 and V2? (An explanation of why the answer is what it is would be appreciated)

What if the polarity of the source were switched around? Would this change the answers for V1 and V2?

Anyone else that would like to add input as well, that'd be great... the more the better,
Thanks

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#### t_n_k

Joined Mar 6, 2009
5,455
Using voltage divider method ...

V1 = V*R1(R1+R2)
V2 = -V*R2/(R1+R2)

The sign obtained in each case is found by considering the current direction in the loop and noting the reference (or ground) point in relation to each of the nodes.

V1-V2 = V12 = V

Another way to view this is as follows:

1. Find the current in the loop, I = V/(R1+R2)

2. Find the voltage across R1,
VR1 = I*R1 = V/(R1+R2)*R1 = V*R1/(R1+R2)

3. The voltage across R1 is the voltage at Node 1 relative to ground

4. Hence V1 = V*R1/(R1+R2) [same polarity as VR1]

5. The voltage across R2,
VR2 = I*R2 = V/(R1+R2)*R2 = V*R2/(R1+R2)

6. Given the current direction the polarity of the voltage across R2 will
be positive at the bottom (ground) and negative at the top (Node 2)

7. Hence V2 = -VR2 = -V*R2/(R1+R2)

8. Alternatively for V2, one may observe that
V2 = V1-V = V*R1/(R1+R2)-V = -V*R2/(R1+R2) [after rearranging]
which agrees with the other approaches

If the source polarity was reversed, the magnitudes would be the same but the polarities of V1 & V2 would be reversed.

It would be worth your while to note the voltage polarities on your sketch and this will hopefully aid with understanding how the node voltage polarities are determined.

#### ihaveaquestion

Joined May 1, 2009
314
Very good explanation tnk, thanks...

I guess this is why I was confusing myself...

I forgot the basics, that parallel terminals have the same voltages while in series have different...

for example in the picture attached, V1=V2=V3 (this is why I wanted to say that the two voltages marked at the top right and top left in my last post were V/-V when in fact they weren't)...

in this same picture attached, again V1=V2=V3 (parallel), and V4 is indeed different because of the voltage drop across that resistor since it's in series.... compared to last example where the source and two resistors are in series...so there are voltage drops across both resistors and they are therefore not just V and -V...

right?

Thanks