^{simulate this circuit – Schematic created using CircuitLab}

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**.

## Best Answer

If you want to prove it the laborious way and ignore the obvious solution just imagine that Vsupply was in fact two identical supplies each independently feeding R2 and R1 respectively.

Then turn those voltage sources into current sources (Norton's and Thevenin's theorums) each respectively in parallel with R2 and R1. Now you end up with: -

Now it's a lot easier to solve. Can you take it from here?