Figure shows the fundamental diagram of Wheatstone bridge. The bridge has four resistive arms, together with a source of emf (a battery) and a galvanometer. The current through the galvanometer depends on the potential difference between the point `c` and `d`. The bridge is said to be balanced when the potential difference across the galvanometer is `0 V` so that there is no current through the galvanometer. This condition occurs when the potential difference from point `c` to point `a`, equals the potential difference from point d to point `a`; or by referring to the other battery terminal, when the voltage drop from other point `c` to point `b` equals the voltage drop from point `d` to point `b`. Hence, the bridge is balanced when

`I_1R_1=I_2R_2`...............(i)

if the galvanometer current is zero, the following conditions also exist:

`I_1=I_3=epsilon/(R_1+R_3)`..............(ii)

and `I_2=I_4 = epsilon/(R_2+R_4)`...........(iii)

Combining equations `(i)`, `(ii)` and `(iii)` and simplifying, we obtain

`(R_1)/(R_1+R_3)=R_2/(R_2+R_4)`................(iv)

from which we get

`R_1R_4=R_2R_3`...............(v)

Equation `(v)` is the well known expression for balance of the Wheatstone bridge. If three of the resistances have known values. The fourth may be determined from equation `(v)`. Hence, if `R_4` is the unknown resistor, its resistance can be expressed in
terms of remaining resistors

`R_4=R_3 R_2/R_1`.................(vi)

Resistance `R_3` is called the standard arm of the bridge and resistors `R_2` and `R_1` are called the ratio arms.