Question

How do you simplify the Wheatstone bridge?

Answer 1

Step 1: Diagram of a Wheatstone bridge

As per the diagram, the current $I$ is from the battery with a potential difference $V$ is bifurcated into equal currents $I_1$ and $I_2$ at $A$. The current through the galvanometer will be $0\mathrm{A}$.

Step 2: Formation of the equation

As per the balanced condition of a Wheatstone bridge, the current through the Galvanometer must be $0$.

In the balanced condition,

$V_1=V_2$

By Ohm's law, the relation will be,

$I_1{R}_1=I_2{R}_2\dots \left(1\right)$

To find the value of currents $I_1$ and $I_2$, apply Ohm's law again, then

$I_1=\frac{V}{R_1+R_3}$

$I_2=\frac{V}{R_2+R_4}$

Step 3: Relation between the resistances

Put the value of $I_1$ and $I_2$ in equation $\left(1\right)$,

$\begin{array}{rcl}\therefore \frac{V{R}_1}{R_1+R_3}& =& \frac{V{R}_2}{R_2+R_4}\\ & \Rightarrow & \frac{R_1}{R_1+R_3}=\frac{R_2}{R_2+R_4}\\ & \Rightarrow & \frac{R_1}{R_1}+\frac{R_1}{R_3}=\frac{R_2}{R_2}+\frac{R_2}{R_4}\end{array}$

Simplify further,

$\begin{array}{rcl}& \Rightarrow & 1+\frac{R_1}{R_3}=1+\frac{R_2}{R_4}\\ & \Rightarrow & \frac{R_1}{R_3}=\frac{R_2}{R_4}\\ & \Rightarrow & R_1{R}_4=R_2{R}_3\end{array}$

Hence, the Wheatstone bridge can be simplified in this way.