# Superposition Principle / Double Current Divider

#### fiolvb

Joined Jan 18, 2016
2
Hey guys, I'm trying to figure out how to use the superposition principle in this circuit to get the current through R4.. I know the basics of superposition
So I first open-circuited the current source and ended up with a total resistance that couldn't help me find current i4...
Then I also tried to short-circuit the voltage source ending up with a total resistance given by r1 in parallel with a series-connection between r2 in parallel with r4 and r3... if this is right, what should I do to get current I4?

#### Jony130

Joined Feb 17, 2009
5,212
Have you try redrawn your circuit.
For example if we remove the current source we get this circuit:

And finding I4 for this circuit is a trivial task.

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#### fiolvb

Joined Jan 18, 2016
2
Have you try redrawn your circuit.
For example if we remove the current source we get this circuit:
View attachment 98961
And finding I4 for this circuit is a trivial task.
Yeah from that point I would join the other three resistors in a parallel and then series connection into one piece, so I would have R_123 in series with R4, but I feel that's wrong, isn't it?
Or could I apply KCL to that node? How do I get the other 2 currents then?

#### Jony130

Joined Feb 17, 2009
5,212
I would join the other three resistors in a parallel and then series connection into one piece, so I would have R_123 in series with R4,
No, the this is the right approach.

#### WBahn

Joined Mar 31, 2012
26,398
Yeah from that point I would join the other three resistors in a parallel and then series connection into one piece, so I would have R_123 in series with R4, but I feel that's wrong, isn't it?
But those three resistors are not in parallel. R1 and R3 are in series and that combination is in parallel with R2.

Or could I apply KCL to that node? How do I get the other 2 currents then?
You can use any circuit analysis technique you want (that is allowed by the instructor). But reducing the circuit to an effective resistance such that the current through that resistance also happens to be the current that you are seeking is probably the easiest way to go.