The Wheatstone bridge is a type of circuit usually used to measure resistance with high precision. The circuit is represented on the following scheme:

Using Kirchhoff’s law we can find that if the current flowing from point B to point D is equal to zero (if there is no current flowing between the two points) that the next formula is true for the resistances:

\[\frac{R_1}{R_2} = \frac{R_4}{R_5}\]

This also works in reverse order, if the resistors fulfill the formula above, no current will flow between points B and D, no matter the resistance or apparatus put in between them.

This gives us the next statement:

\[\frac{R_1}{R_2} = \frac{R_4}{R_5} ⟺ I_{BD} = 0\]

This means that, when observing current or resistance of the whole circuit, we can eliminate the current between points B and D because no current is flowing, and we can neglect the resistance between the 2 points because it won’t contribute to the total resistance of the circuit. Because of this, when redrawing the circuit for simplicity, we can redraw the part between the 2 points where current isn’t flowing (points B and D in our case) as a wire without any apparatus in between them, or we can just redraw the bridge without the wire between the 2 points, giving us the next, much simpler circuit:

How is the Wheatstone bridge used to measure resistance?

By putting a galvanometer that measures current between points B and D, we know that when the galvanometer measures a current of 0, the already mentioned formula is satisfied. Because of this, if we know the resistances of 3 of the resistors, we can easily find the resistance of the fourth one that we are trying to find.

Find x if the bridge is balanced.

Solution:

When the bridge is balanced, it means there is no current flowing through the galvanometer, so we get the next equation:

\(\frac{R}{P}= \frac{S}{Q}\)

\(R=\frac{SP}{Q}\)

\(x + 400 = \frac{1000 \times 500}{800}\)

\(x + 400 = 625\)

\(x = 625 - 400\)

\(x = 225 \, \Omega\)

Written by Nemanja Maslak