Hi. I'm struggling on what I thought looked like a simple problem. I need the current at i1. I actually know that the answer is 1.25A, but I can't come up with it. I'm apparently getting off while trying to reduce the parallel and series resistors to an equivalent with hopes of getting the current division in the first loop.

(this problem comes from an ancient edition of Nilsson's Electric Circuits, 2nd edition, btw)

Thanks

 

Unfortunately there are no immediate opportunities to do any series or parallel resistor reductions in this circuit. You could bring out the heavy machinery and use a Y-Delta or Delta-Y transformation to get things moving, but that seems like a lot of effort.

Might as well just go ahead and employ Kirchhoff's laws to solve the circuit.

Hint: A resistor in series with a current supply has no effect on the rest of the circuit. If the value of the current source is I, then there will be I passing through that branch no matter what the value of that resistor is. You can safely remove that resistor entirely from the circuit for purposes of analysis.

 

Hello andrewintejas

That's probably not as complex as painted by gneill, of course if you study thoroughly the following videos.

http://www.youtube.com/watch?v=_kJURIUPllw
http://www.youtube.com/watch?v=4mmvKpueSfI
http://www.youtube.com/watch?v=qWi4D50bsqA

 

Hi,

You could use simple Nodal Analysis also. You could apply a voltage across the bridge part and solve for the upper central node voltage v1 and the lower central node voltage v2. You could then calculate the current through the network with a 1v voltage source and then solve for the equivalent resistance. You would then multiply times the current source and use that voltage and the solution for v2 to calculate I1.

Doing this we get v1=2v/3 and v2=v/2, and setting v=1 we get a left side current of 1/6 amps, so at 1 volt that means the bridge network is 6 ohms. Multiplying this by 2.5 we get 15 volts, and since v2=v/2 we get 7.5 volts for v2, and 7.5/6=1.25 amps.

 

Yeah, you don't want to do any resistor simplification/reduction.

What you want to do is use Node-Voltage or Mesh-Current methodes. Pick one and use it.

If you still want to do the resistor simplification/reduction, you have to use Delta to Wye resistor transformation. This will rearange the resistors so that you can do that thing that you want to do. However, it is kinda waste of time since you don't need it.

 

Yeah, you don't want to do any resistor simplification/reduction.

What you want to do is use Node-Voltage or Mesh-Current methodes. Pick one and use it.

If you still want to do the resistor simplification/reduction, you have to use Delta to Wye resistor transformation. This will rearange the resistors so that you can do that thing that you want to do. However, it is kinda waste of time since you don't need it.

Hi,

You dont really have to use delta to wye or vice versa you can still use good ol' nodal to get the resistor simplification. It leads to 6 ohms which then makes it easier to solve.

 

Hi,

You dont really have to use delta to wye or vice versa you can still use good ol' nodal to get the resistor simplification. It leads to 6 ohms which then makes it easier to solve.

I would just use mesh current methode, setup the three equations, plug them into calculator and collect the answers. Then use the two of the three mesh currents to find I1.

I would not bother doing anything with resistors. There is no need. But if you want to... then do that thing that you want to do.

 

Thanks so much everyone! I missed the obvious delta-Y on the resistor piece, which easily converted to 6 ohms once setup as such. I was also able to solve for i1 without any of the resistor business per your advice. Just simple KCL, KVL

KVL:
4a + 10b - 6c = 0
10d -5e -10b = 0

KCL:
a + c = 2.5
a = b + d
d + e = 2.5

where c = i1 and d= i2
and which correctly yields i1 = 1.25 and i2 = 1

That's what I get for working circuit problems at 3am...