Finding voltage and current in a somewhat unusual circuit

[Bassalisk]

The circuit is essentially an unbalanced Wheatstone bridge. A nifty technique for solving it is to break the two vertical resistor combinations apart as shown in the attachment.

The top half of the image shows how to break the two halves of the bridge and apply identical 50 volt sources.

The bottom half of the image shows the Thevenin equivalent of each half. Since we have two voltage sources of 100/9 volts and 50/7 volts respectively, with corresponding Thevenin resistances of 140/9 ohms and 30/7 ohms, we can find the current in Rwire by applying a voltage of (100/9-50/7) volts to a resistance of (140/9+30/7+Rwire) ohms and solving for the current.

The current is given by (100/9-50/7)/(140/9+30/7+Rwire) = (250/63)/(1250/63 + R) = 250/(63*R+1250)

We can even get a value for current if Rwire is 1 micro-ohm or even zero. In case Rwire is zero, the current is 1/5 = .2 amp.

Yes I agree, I was going to go after Thevenin but since mate told us that he is beginner i dropped it. But well done!

 

[MATLABdude]

There's an even simpler and more germane illustration of the "current within node" problem. [URL]http://upload.wikimedia.org/wikipedia/commons/4/4f/Current_division_example.svg[/URL]
From the Wikipedia article on current divider:
http://en.wikipedia.org/wiki/Current_divider

When you change the current source for a voltage source, you can see that all the resistors and the source share the same two common nodes. However, it's plain to see that current must move within the two common nodes to supply current to each of the resistors, even when you assume the wire has zero resistance.

As others have said, an ideal node just implies that everything connected to it is at the same voltage / potential, it says nothing about the currents within. Don't be too non-plussed though, 0A was my answer when I first glanced at it, too!