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How to prove that equivalent resistance of any passive network is always lower if we add a resistor between arbitrary two nodes?

Note that this is not necessarily a parallel circuit, 2 nodes that we connect with a resistor are not the same 2 nodes between we want to measure equivalent resistance but completely arbitrary 2 nodes in passive network.

So imagine a circuit with 4 access points: A, B, C and D. We want to measure equivalent resistance between A and B. How can you prove that Rab will be less if we add a resistor between C and D?

I tried to search the web but didn't have any success.

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To clarify, I am interested in equivalent DC resistance in an arbitrary network made of resistors only. How can we prove that the resistance Rab is not higher if we connect nodes C and D with any resistor? –  Serge Dec 13 '12 at 12:46
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2 Answers

Instead of trying to prove this conjecture, you might instead look for counter-examples. I think there are passive networks in which adding a resistor between C and D leaves the equivalent resistance between A and B unchanged. A trivial counter-example is a network in which the only connections of node C are to D; there are others as well.

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Ok, I'll rephrase, lower or equal. –  Serge Dec 13 '12 at 9:41
    
@Serge: Well, you might try using the two-port open-circuit impedance matrix representation of the circuit, writing down the formula for the port-1 input resistance when port-2 is terminated by a resistor, and then use passivity and reciprocity to establish contraints on the various z-elements. –  Art Brown Dec 13 '12 at 19:43
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A balanced Wheatstone bridge is a nontrivial example. –  Emilio Pisanty Mar 14 '13 at 0:46
    
@EmilioPisanty: yes that's a good one. –  Art Brown Mar 14 '13 at 2:11
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Convert to the equivalent two-port representation with 4 nodes A,B,C,D and 6 resistors, one resistor between each node. Note the resistor directly between A and B is not important to this question. So you have a diamond with your test ports A and B on top and bottom, and a cross-resistance between C and D.

Figure out the equivalent resistance of this 5-resistor network. I'll use the extra element theorem. It is: $$R_{eq} = R^\infty_{in} \frac{R_{cd} + R^0_x}{R_{cd} + R^\infty_x}$$ where $$R^\infty_{in} = (R_{ac} + R_{bc})\|(R_{ad} + R_{bd})$$ $$R^0_x = (R_{ac} \| R_{bc}) + (R_{ad} \| R_{bd})$$ $$R^\infty_x = (R_{ac} + R_{ad}) \| (R_{bc} + R_{bd})$$

$R_{cd}$ only appears in the first equation, and so $R_{eq}$ is an increasing function of $R_{cd}$ provided $R^0_x < R^\infty_x$. If $R^0_x = R^\infty_x$ then $R_{eq}$ is independant of $R_{cd}$. (That is also the most general counterexample to the original question.)

Prove algebraically that $R^0_x \le R^\infty_x$.

Even though you are only interested in DC resistance, it is true for general passive impedance.

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