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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

3 Answers 3

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

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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An old question, but it got me thinking. And I did get an answer, so here it is...

We have a circuit, with $n$ nodes, numbered $0\ldots n-1$. There are also $b$ branches, numbered $0\ldots b-1$. Node number $0$ ($n_0$) is chosen to be the reference, or ground node. Node number $1$ ($n_1$) is kept $1 \mathrm{V}$ by a voltage source connected from $n_0$ to $n_1$. This is branch $0$ ($b_0$). The current on $b_0$ is defined to be flowing from $n_0$ to $n_1$, making $v_0$ (voltage of branch $0$) $-1\mathrm{V}$ by convention.

The equivalent resistance of the circuit as seen from nodes $n_1 - n_0$ is defined to be $1V / i_0$.

Now, in addition to $b_0$ consider another branch $b_1$, which is initially "not present", that is, initially $i_1 = 0$.

Here, we invoke Tellegen's theorem, which states the following:

Let $I_k$ be a set of branch currents on a circuit satisfying KCL.

Let $V_k$ be a set of branch voltages on a circuit satisfying KVL.

Then, for any such sets,

$\sum I_k\cdot V_k = 0$

Note that voltages and currents need only satisfy KCL and KVL, they need not be related to each other!

Now, going back to our setup, we have two sets of currents and voltages.

First, with $b_1$ not present.

$i_k$, $v_k$

$\sum i_k \cdot v_k = 0$

Expanding the special branches we have:

$v_0 i_0 + v_1 i_1 + \sum\limits_{k=2}^{b-1} v_k i_k = 0$

Since $v_0 = -1$ and $i_1 = 0$ we have:

$i_0 = \sum\limits_{k=2}^{b-1} v_k i_k$

When $b_1$ is inserted with a finite resistor, all currents and voltages are different. So we get:

$i'_0 = v'_1 i'_1 + \sum\limits_{k=2}^{b-1} v'_k i'_k$

Unfortunately, this takes us nowhere. But, there is a different approach: Use Tellegen's theorem, but with voltages and currents crossed between the two situations. This is perfectly valid:

$i_0 = \sum\limits_{k=2}^{b-1} v'_k i_k$

$i'_0 = v_1 i'_1 + \sum\limits_{k=2}^{b-1} v_k i'_k$

Up to here, everything is general. Now, we use the fact that, apart from $b_0$ which contains a voltage source, every element is a resistor:

$v_k = R_k \cdot i_k$

and

$v'_k = R_k \cdot i'_k$

Substituting theses for the voltages, the equations become:

$i_0 = \sum\limits_{k=2}^{b-1} R_k i'_k i_k$

$i'_0 = v_1 i'_1 + \sum\limits_{k=2}^{b-1} R_k i_k i'_k$

Note that the sums are exactly the same! Subtract the two equations to obtain:

$i'_0 - i_0 = v_1 i'_1$

Now, $v_1$ is the voltage where we are going to connect the new resistor before we have connected it. $i'_1$ is the current going through it after we have connected it. The product is definitely positive. This can be proven by considering the rest of the circuit as seen by the new resistor as a Thévénin equivalent circuit; the later current must be in the same direction as the initial voltage.

Since $R_{eq} = \frac{1V}{i_0}$ and $R'_{eq} = \frac{1V}{i_1}$, it follows that since $i'_0 > i_0$ then $R'_{eq} < R_{eq}$.

Well, we even know exactly by how much!

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