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schematic

We know that in this Wheatstone bridge if, $\bf{I_3 = 0}$, it can be derived that $\bf{\frac{R_1}{R_5}=\frac{R_2}{R_4}}$ . But could we prove it backwards, mathematicaly, that - if in such a circuit $\bf{\frac{R_1}{R_5}=\frac{R_2}{R_4}}$, then it will be also true that $\bf{I_3 = 0}$?

I am asking this question because I've seen in the case of solving such circuit problems this argument being made using the reference of Wheatstone bridge, that as $\bf{\frac{R_1}{R_5}=\frac{R_2}{R_4}}$, so the equivalent $\bf{I_3}$ would be 0.

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  • $\begingroup$ You're dealing with a linear circuit, so yes, the two statements are equivalent. $\endgroup$
    – QtizedQ
    Jul 17, 2017 at 10:48
  • $\begingroup$ @QtizedQ Sorry, I don't think I understand what a linear circuit is. But as you said, in this circuit the two statements are equivalent; so, can we mathematicaly prove the other? $\endgroup$
    – Tiash
    Jul 17, 2017 at 10:57
  • $\begingroup$ I'll write up a proof when I get to work. But if you assume those ratios of resistances and solve the circuit, what potential would you get out? $\endgroup$
    – QtizedQ
    Jul 17, 2017 at 11:16

3 Answers 3

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Remove $R_3$ and determine the open circuit voltage $V_3^\mathrm{oc}$ across the diagonal of the bridge (positive on the left node),

$$\begin{align}V_3^\mathrm{oc} &= \left(\frac{R_4}{R_2+R_4}-\frac{R_5}{R_5+R_1}\right)V \\ &= \frac{R_1R_4-R_2R_5}{(R_2+R_4)(R_1+R_5)}V \\ &= \frac{R_1/R_5-R_2/R_4}{(1+R_2/R_4)(1+R_1/R_5)}V \end{align}$$

From Thévenin's theorem, the current $I_3$ is proportional to $V_3^\mathrm{oc}$, and from the above equation this voltage is zero if and only if the numerator is zero, that is,

$$\frac{R_1}{R_5} = \frac{R_2}{R_4}.$$

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Let $V_A$ be the voltage at the left node of $R_3$, and let $V_B$ be the node voltage to the right of $R_3$. Let $V_3 = V_A-V_B$ be the voltage across $R_3$. Now, we want to solve for $V_3$ as a function of the source voltage $V_S$ and the resistors $R_i$. We have two independent unknowns to solve for ($V_A$ and $V_3$ full determine $V_B$), so we can write down KCL at the $A$ and $B$ nodes in terms of $V_A,V_3$ and constants to determine the two voltages.

First, we have the KCL at the left node:

$$\frac{V_S-V_A}{R_2} = \frac{V_3}{R_3} + \frac{V_A}{R_4}$$

And then the right node, using $V_B=V_A-V_3$:

$$\frac{V_S-(V_A-V_3)}{R_1} + \frac{V_3}{R_3} = \frac{V_A-V_3}{R_5}$$

Solving these equations using your favorite technique, we find

$$V_3 = \frac{R_3V_S(R_1R_4-R_2R_5)}{...}$$

where I have omitted the denominator, which is a sum of products of three-resistor combinations.

This also tells us that $I_3 = \frac{V_3}{R_3} = \frac{V_S(R_1R_4-R_2R_4)}{...}$

Note that the numerator tells us everything we need. Taking $R_1R_4=R_2R_5$, or similarly $\frac{R_1}{R_5} = \frac{R_2}{R_4}$ assuming nonzero resistors, we see that this makes the numerator zero, which means that the voltage $I_3$ is also zero.

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  • $\begingroup$ Sorry to bother you again. But in the first step of your proof, you used the relation of voltage at a potential divider to the total input voltage and the resistors, that ( $V_{out} = \frac{R2}{R1+R2}V_{in}$}. But this relation requires the current through the potential divider to be zero. link So, in order to make the first statement in your proof true, don't we again have to make the assumption that current flowing through the potential divider is zero, first? $\endgroup$
    – Tiash
    Jul 22, 2017 at 3:28
  • $\begingroup$ I'll redo it now that you've revised the circuit so the central element is a resistor. $\endgroup$
    – QtizedQ
    Jul 22, 2017 at 12:59
  • $\begingroup$ I have updated my answer assuming that the central element is a resistor and derived the result that $I_3$ is zero if the resistors obey that specific ratio. $\endgroup$
    – QtizedQ
    Jul 24, 2017 at 13:50
  • $\begingroup$ Are you sure that solving the two equations gives that in the numerator? How did you get that? $\endgroup$
    – Tiash
    Jun 8, 2018 at 16:07
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The easiest way to picture this is to model the circuit with $R_3$ removed. Then each side of the bridge is just a simple voltage divider, with voltages at the center of $$V\frac{R_4}{R_2+R_4}=V\left(1+\frac{R_2}{R_4}\right)^{-1}$$ and $$V\frac{R_5}{R_1+R_5}=V\left(1+\frac{R_1}{R_5}\right)^{-1}$$ respectively.

These two voltages are equal if (and only if, assuming V is non-zero) $\frac{R_2}{R_4}=\frac{R_1}{R_5}$.

Since these voltages are equal, adding a resistor (or a wire/capacitor/inductor: basically anything except a current or voltage source) between these two nodes does not change anything: no current flows through the additional circuit element.

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