Parallel or in series?Superposition problems

[esmeco]

I'm having a bit of a difficulty determining some of the resistors in parallel and in series with each other...

In the image in attachment when we use the Superposition theorem and remove the 1ma current source and substitute it by an open circuit,how are
R1 and R3 related to each other,series or parallel?
Also,because of the open circuit shouldn't be R1 removed,since the current passing through it equals 0?
And what about R2 and R3 when we remove the 3ma current source?In my perspective I think that it would be equal to removing them,since there won't be any current flowing throw them since the path where the current flowed through the + side of the voltage source to the - side has been cut off.

Any help on this is very welcomed!

 

[ranger]

In the image in attachment when we use the Superposition theorem and remove the 1ma current source and substitute it by an open circuit,how are
R1 and R3 related to each other,series or parallel?

With the 1mA current source removed, does there exit a complete path between R1 and R3?

Also,because of the open circuit shouldn't be R1 removed,since the current passing through it equals 0?

Bingo! This should help you with the first question.

And what about R2 and R3 when we remove the 3ma current source?In my perspective I think that it would be equal to removing them,since there won't be any current flowing throw them since the path where the current flowed through the + side of the voltage source to the - side has been cut off.

Well R2 is removed, but R3 is still in the loop of R1 and 1mA.

BTW, in superposition, you have to consider one source at a time by replacing the other two with their ideal internal resistance. It sounds to be like you're only removing one of them.

 

[esmeco]

So,using what you've said would this be correct:

->Removing I1

Suppose the current passing through R3 is I3

I3=I2
V3=R3xI2 <=> v3=1x3=3v

->Removing I2

I3=I1
V3=R3xI1 <=> v3=1x1=1v

->Sort circuiting the voltage source

I1+I2=I3
V3=1x4=4v

So the contribution of the sources with result in:

Vt=4+1+3=8v

Is this right?

 

[ranger]

So,using what you've said would this be correct:

->Removing I1

Suppose the current passing through R3 is I3

I3=I2
V3=R3xI2 <=> v3=1x3=3v

This is good. But remember what I said above - only treat one source at a time. So it should have read "Removing I1 and shorting V1"

->Removing I2

I3=I1
V3=R3xI1 <=> v3=1x1=1v

Yup this is current also. But once again it should have read "Removing I2 and shorting V1"

->Sort circuiting the voltage source

I1+I2=I3
V3=1x4=4v

Not sure I follow you here. You've already established the current through R3 due to I1 and I2, so why are you shorting the voltage source? You have to find the current due to V1 by removing the I1 and I2.

 

[esmeco]

Well I short circuited the voltage source because I thought that we should do one source at a time(first open circuit I1 and the voltage source and I2 remained,then open circuit I2 and then the voltage source and I1 remained,and,at last,short circuit the voltage source while I1 and I2 remained)?Shouldn't it be this way?

 

[ranger]

Well I short circuited the voltage source because I thought that we should do one source at a time(first open circuit I1 and the voltage source and I2 remained,then open circuit I2 and then the voltage source and I1 remained,and,at last,short circuit the voltage source while I1 and I2 remained)?Shouldn't it be this way?

You misunderstood when I said one source at a time. By this I mean only one source should be present in the circuit at any given time. This is the principle of the superposition theorem. When we have a circuit with multiple sources and we wish to find the resultant current (or voltage) across a resistor, we must find the contribution of each of the sources [individually with the others replaced by their ideal internal source resistance].

 

[esmeco]

Humm,now I get it...So,if I understood you correctly,if we have,for example,2 voltage sources and 1 current,if we want to determine the contribution of the current source we open circuit the current source and short circuit both voltage sources?

 

[ranger]

To determine the contribution of the current source in that scenario, we would leave the current source and short the two voltage sources.

 

[esmeco]

Ah,ok...just one more thing,to determine the contribution of one of the voltage sources we short circuit one of the voltage source and open circuit the current source,right?If that so,why shouldn't it be done for the circuit on the picture?Shouldn't we also be wanting the contribution of the voltage source?

 

[ranger]

Ah,ok...just one more thing,to determine the contribution of one of the voltage sources we short circuit one of the voltage source and open circuit the current source,right?

For the scenario you described in post #7, yes.

If that so,why shouldn't it be done for the circuit on the picture?Shouldn't we also be wanting the contribution of the voltage source?

We absolutely want to find the contribution (if any) from the voltage source. So go ahead and remove the two current sources and tell us what happens.

 

[esmeco]

Ok,I understand now...The contribution of the voltage source would be zero since there won't be any current flowing through the resistors thanks to the open circuit,so it wouldn't be necessary to calculate the contribution,right?

 

[ranger]

Ok,I understand now...The contribution of the voltage source would be zero since there won't be any current flowing through the resistors thanks to the open circuit,so it wouldn't be necessary to calculate the contribution,right?

Yup, you are right. So what's the voltage (and current) across R3?

 

[esmeco]

Both equal to zero?

 

[ranger]

I meant to ask what the resultant current through R3 as a result of all the sources?

 

[esmeco]

Well it would be equal to sum of the currents contributions (1+3=4ma) and the voltage would also be equal to the sum of the contributions which would equal 4v,I guess...

 

[ranger]

Well it would be equal to sum of the currents contributions (1+3=4ma) and the voltage would also be equal to the sum of the contributions which would equal 4v,I guess...

Yup, you've go it.

 

[esmeco]

Thank you for all the help!:D

 

[ranger]

Sure, anytime.