# Wheatstone bridge

## Wheatstone bridge

For measuring accurately any electrical resistance Wheatstone bridge is widely used. There are two known resistors, one variable resistor and one unknown resistor connected in bridge form as shown below. By adjusting the variable resistor the current through the Galvanometer is made zero. When the current through the galvanometer becomes zero, the ratio of two known resistors is exactly equal to the ratio of adjusted value of variable resistance and the value of unknown resistance. In this way the value of unknown electrical resistance can easily be measured by using a Wheatstone bridge. Wheatstone bridge Theory: The general arrangement of Wheatstone bridge circuit is shown in the figure below. It is a four arms bridge circuit where arm AB, BC, CD and AD are consisting of electrical resistances P, Q, S and R respectively. Among these resistances P and Q are known fixed electrical resistances and these two arms are referred as ratio arms. An accurate and sensitive Galvanometer is connected between the terminals B and D through a switch S₂.

The voltage source of this Wheatstone bridge is connected to the terminals A and C via a switch S₁ as shown. A variable resistor S is connected between point C and D. The potential at point D can be varied by adjusting the value of variable resistor. Suppose current I₁ and current I₂ are flowing through the paths ABC and ADC respectively. If we vary the electrical resistance value of arm CD the value of current I₂ will also be varied as the voltage across A and C is fixed. If we continue to adjust the variable resistance one situation may come when voltage drop across the resistor S that is I₂. S is becomes exactly equal to voltage drop across resistor Q that is I₁.Q. Thus the potential at point B becomes equal to the potential at point D hence potential difference between these two points is zero hence current through galvanometer is nil. Then the deflection in the galvanometer is nil when the switch S₂ is closed.

Now, from Wheatstone bridge Circuit:

Current (I₁) = $$\frac{V}{P\,+\,Q}$$,

Current (I₂) = $$\frac{V}{R\,+\,S}$$,

Now potential of point B in respect of point C is nothing but the voltage drop across the resistor Q and this is $${{I}_{1}}Q\,=\,\frac{VQ}{P+Q}$$ … (1)

Again potential of point D in respect of point C is nothing but the voltage drop across the resistor S and this is $${{I}_{2}}S\,=\,\frac{VS}{R+S}$$ … (2)

Equating, equations (1) and (2) we get,

$$\frac{VQ}{P+Q}\,=\,\frac{VS}{R\,+\,S}$$,

$$\frac{Q}{P\,+\,Q}\,=\,\frac{S}{R\,+\,S}$$,

$$\frac{P+Q}{Q}\,=\,\frac{R+S}{S}$$,

$$\frac{P}{Q}\,+\,1\,=\,\frac{R}{S}\,+\,1$$,

$$\frac{P}{Q}\,=\,\frac{R}{S}$$,

$$R\,=\,S\times \frac{P}{Q}$$.

Here in the above equation, the value of S and P⁄Q are known so value of R can easily be determined. The electrical resistances P and Q of the Wheatstone bridge are made of definite ratio such as 1:1; 10:1 or 100:1 known as ratio arms and S the rheostat arm is made continuously variable from 1 to 1,000Ω or from 1 to 10,000 Ω.