# A Brief on Wheatstone bridge and Its Working

The term “Wheatstone bridge” is also called as Resistance Bridge that is, invented by “Charles Wheatstone”. This bridge circuit is used to calculate the unknown resistance values and as a means of regulating measuring instrument, ammeters, voltmeters, etc. But, the present digital millimeters offer the easiest way to calculate a resistance. In recent days, Wheatstone bridge is used in many applications such as; it can be used with modern op-amps to interface various sensors and transducers to amplifier circuits. This bridge circuit is constructed with two simple serial and parallel resistances in between a voltage supply terminal and ground terminals. When the bridge is balanced, then the ground terminal produces a zero voltage difference between the two parallel branches. A Wheatstone bridge consists of two i/p and two o/p terminals includes of four resistors arranged in a diamond shape.

## Wheatstone Bridge and Its Working

A Wheatstone bridge is widely used to measure the electrical resistance. This circuit is built with two known resistors, one unknown resistor and one variable resistor connected in the form of bridge. When the variable resistor is adjusted, then the current in the galvanometer becomes zero, the ratio of two two unknown resistors is equal to the ratio of value of unknown resistance  and adjusted value of variable resistance. By using a Wheatstone Bridge  the unknown electrical resistance value can easily measure.

### Wheatstone Bridge Circuit Arrangement

The circuit arrangement of  the Wheatstone bridge is shown below.  This circuit is designed with four arms, namely AB, BC, CD & AD and consists of  electrical resistance P, Q, R and S.  Among these four resistances, P and Q are known fixed electrical resistances. A galvanometer is connected between the B & D terminals via an S1 switch. The voltage source is connected to the A &C terminals via a switch S2. A variable resistor ‘S’ is connected between the terminals C & D. The potential at terminal D varies when the value of the variable resistor adjusts. For instance, currents I1 and I2 are flowing through the points ADC and ABC. When the resistance value of arm CD varies, then the I2 current will also vary.

If we tend to adjust the variable resistance one state of affairs could return once when the voltage drop across the resistor S that is I2.S  becomes specifically capable to the voltage drop across resistor Q i.e I1.Q. Thus the potential of the point B becomes equal to the potential of the  point D hence the potential difference b/n these two points is zero hence current through galvanometer is zero. Then the deflection in the galvanometer is zero when the S2 switch is closed.

Wheatstone Bridge Derivation

From the above circuit, currents I1 and I2 are

I1=V/P+Q and I2=V/R+S

Now potential of point B with respect to point C is the voltage drop across the Q transistor, then the equation is

I1Q= VQ/P+Q …………………………..(1)

Potential of point D with respect to C is the voltage drop across the resistor S, then the equation is

I2S=VS/R+S …………………………..(2)

From the above equation 1 and 2 we get,

VQ/P+Q = VS/R+S

`                                 Q/P+Q = S/R+S

P+Q/Q=R+S/S

P/Q+1=R/S+1

P/Q=R/S

R=SxP/Q

Here in the above equation, the value of P/Q and S are known, so R value can easily be determined.

The electrical resistances of Wheatstone bridge such as P and Q are made of definite ratio, they are 1:1; 10:1 (or) 100:1 known as ratio arms and the rheostat arm S is made always variable from 1-1,000 ohms or from 1-10,000 ohms

### Example of Wheatstone Bridge

The following circuit is an unbalanced Wheatstone bridge, calculate the o/p voltage across C and D points and the value of the resistor R4 is required to balance the bridge circuit.

The first series arm in the above circuit is ACB
Vc= (R2/(R1+R2)) X Vs
R2=120ohms, R1=80 ohms, Vs=100
Substitute these values in the above equation
Vc= (120/(80+120)) X 100
= 60 volts
The second series arm in the above circuit is ADB

VD = R4/(R3+R4) X Vs

VD= 160/ (480+160) X 100
=25 Volts
The voltage across points C & D is given as
Vout= VC-VD
Vout= 60-25 = 35 volts.
The value of  R4 resistor is  required to balance the Wheatstone bridge bridge is given as:
R4= R2 R3/R1
120X480/ 80
720 ohms.

So, finally we can conclude that, the Wheatstone bridge has two i/p & two o/p terminals namely A & B, C& D. When the above circuit is balanced, the voltage across the o/p terminals is zero volts. When the Wheatstone bridge is unbalanced, the o/p voltage may be either +ve or –ve depending upon the unbalance direction.

#### Application of Wheatstone Bridge

The application of Wheatstone bridge is light detector using Wheatstone bridge circuit