#### Brandon_1539537396

Joined Oct 14, 2018
5
Ok so I'm lucky enough to have an inaccessible substandard PhD candidate as a professor and he's terrible at helping conceptualize the material. My issue is that I am having trouble discerning which resistors play a part in the superposition problem. Parallel resistors to an open circuit are supposed to effectively removed from the equation according to him, but if that's the case, then there would be no current at all through them and this entire question is futile as it wouldn't require superposition at all. I'm also at a loss at to which resistors are in parallel and/or in series respective to their source current and voltage.
Basically I'm at a loss at to where to start, and the internet wanted \$70 to answer this one question. Any help and guidance would be incredibly appreciated.

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#### ericgibbs

Joined Jan 29, 2010
16,424
hi Brandon,
Welcome to AAC.
As this is homework we would like to see your attempt at solving it, then we can point out any errors.
E

#### Brandon_1539537396

Joined Oct 14, 2018
5
Here's what I came up with. If I'm right, please confirm. If I'm off (as I suspect) please let me know why. Thanks.

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#### The Electrician

Joined Oct 9, 2007
2,912
First, the left hand pane of your work. Note that R4 has 12 volts across it, regardless of whether I1 is present or not. Ohm's law says that when you have a voltage V across a resistor R, the current in that resistor is V/R. What then is the current through R4?

With I1 missing, you have three resistors (R1, R2, R3) in series, and that series combination also has 12 volts across it. If you have three resistors in series the current in each of them is the same, so the current in R1 is equal to the current in the series combination with 12 volts applied to the combination.

Second, the right hand pane of your work. When you replace V1 with a short, that short also shorts out R4, so R4 is no longer in play. Recalculate the currents with R4 and V1 both replaced with a short.

#### Brandon_1539537396

Joined Oct 14, 2018
5
Ok but the voltage is the same across parallel resistors, but split between series resistors, correct? And since botit's both, would the voltage be reduced? Also, why is I1 missing?
Lastly, when something is replaced by a short, how can you decide which resistors are shorted out also?

#### Brandon_1539537396

Joined Oct 14, 2018
5
That middle sentence should read: And since it's both (series and parallel) wouldn't the voltage be reduced?
Sorry, sucks working full time and scrambling to finish this simultaneously...

#### MrAl

Joined Jun 17, 2014
9,556
Hello,

You can not combine the 1k and 4k to make 5k when you kill the current source.
Look at that again.
The voltage source is connected across the 4k that's why you cant do that.
You can however combine the 1k in another way...look for that.
I did not look at anything else yet i thought i would tell you that first and see if you could rework the problem now.

#### WBahn

Joined Mar 31, 2012
27,486
View attachment 161536 Here's what I came up with. If I'm right, please confirm. If I'm off (as I suspect) please let me know why. Thanks.

Taking the first part of your work:

With the current source removed, the two 2 kΩ resistors are, indeed, in series. But the other two are not. To be in series, two components have to have the same current in them. In this case, current coming out of the bottom resistor going to the right could either go up through the righthand resistor or it could go up through the voltage source, thus they are not in series.

But the bottom resistor IS in series with the other two 2 kΩ resistors since whatever current is flowing in one must flow in all three.

#### Brandon_1539537396

Joined Oct 14, 2018
5
So the current is not dependent on directionality? And the first responder intonated that R1 was affected by the open circuit removal. I'm having trouble mashing the responses into something actionable. So with the open circuit, does R1 still interact with the 12V source? And I'm failing to understand the significance of the 4K resistor being connected "across" the voltage source.

#### WBahn

Joined Mar 31, 2012
27,486
So the current is not dependent on directionality?
What do you mean by this? Please be specific.

And the first responder intonated that R1 was affected by the open circuit removal.
So?

I'm having trouble mashing the responses into something actionable.
In what way? Be specific -- we are NOT mind readers!

So with the open circuit, does R1 still interact with the 12V source?
Sure. Why wouldn't it? I don't see where anyone has said any different.

Draw the circuit with the current source replaced by an open circuit and determine the currents (including direction) through each component. If you can't do that in a circuit having a single source, then you aren't ready to tackle circuits with multiple sources and, instead, need to take a step back and improve those skills first.

And I'm failing to understand the significance of the 4K resistor being connected "across" the voltage source.
What is the voltage across the 4 kΩ resistor?

Do you need to know ANYTHING about the rest of the circuit to answer that question?

If not, and if you know the voltage across the 4 kΩ resistor, do you know the current flowing through that resistor?

Do you need to know ANYTHING about the rest of the circuit to answer that question?

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#### MrAl

Joined Jun 17, 2014
9,556
So the current is not dependent on directionality? And the first responder intonated that R1 was affected by the open circuit removal. I'm having trouble mashing the responses into something actionable. So with the open circuit, does R1 still interact with the 12V source? And I'm failing to understand the significance of the 4K resistor being connected "across" the voltage source.
Hello again,

Here is an illustration of what you would need to get the circuit you got on the right.
Note the voltage source of 12v would have to be connected differently to get that.
It is connected differently in part B while part A is the original drawing.
See if you can tell the difference now.

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