Calculating Current for Two-port Resistor Network

AI Thread Summary
The discussion revolves around calculating the current ratio in a two-port resistor network with specified resistances and a load resistor. A user struggles to model the network correctly and finds the expected current ratio to be 5, with 83.3 mA from the supply and 16.6 mA through the load resistor. Clarifications about equivalent resistances in parallel are discussed, emphasizing the complexity of the problem. The conversation also touches on the potential for multiple answers if the question were rephrased regarding equivalent resistance. Overall, the participants seek to understand the nuances of the problem and the implications of the resistor configurations.
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Homework Statement



A network of resistors has a pair of input terminals AB connected to a dc supply and a pair of output terminals CD connected to a load resistor of 120 ohm. The resistances of the network are AC=BD =180 ohm, and AD=BC=80 ohm. Find the ratio of the current in the load resistor to that taken from the supply.

Homework Equations



Kirchoff's Rules

The Attempt at a Solution



I have tried to figure out a model network having the same resistance describe in the question but failed. Several other attempts are also in vain. The model answer is 5.
 
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The difficulty is in how the question is asked.

If it was drawn like this, could you solve it?

[PLAIN]http://dl.dropbox.com/u/4222062/bridge2.bmp

Assume a voltage that suits you and then work out the currents for that voltage.

You only need the ratio between currents, so this is unaffected by the voltage you choose.
 
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Thanks very much for your idea, vk6kro.

But are you sure the network you draw satisfies the condition given in the question?

Consider the equivalent resistance across AD ( with the 120 ohm resistor disconnected) , it isn't simply equal to 80 ohms, because there is a second branch ( consist of the other three resistors ).

Or have I missed out something... ?

Thanks anyway for your precious time and effort.

Regards
Steven
 
I hadn't thought of those as being the equivalent resistances with the other resistors in parallel. That would be vicious.

Fortunately the person who wrote the question didn't either, because using the above diagram I get 5 times as much current coming out of the battery as flows in the 120 ohm resistor.
83.3 mA from the battery and 16.6 mA in the 120 ohm resistor.

This was the answer you were supposed to get.
 
Okay, so seems that I've misunderstood the question...

Just curious, what if I change the question so that the resistances in the question refers to equivalent resistance? How can I solve this more complicated problem? Or, is it solvable?

Thanks very much, your reply is very helpful.
 
I don't know if you could always solve it if the problem was meant the other way.

I suspect you could get multiple possible answers even if you could solve it.

I think 100% of teachers are genuinely trying to help their students and you would never get a question like that.
 

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