Wheatstone Bridge

The Wheatstone bridge is the setup that is used for measuring the unknown resistance. Wheatstone bridge consists of four arms known as resistors and the ratio of the two of these resistors is kept at a fixed value. The remaining two arms are balanced, one of them is an unknown resistor while the other resistance of the other arm can be varied. 

The unknown resistance is then computed through the balancing or null condition. The circuit of the Wheatstone bridge gives a precise measurement of the resistance. The numerous adaptations of the Wheatstone bridge are utilised for the AC circuits. Some of the instruments on the Wheatstone bridge principle are Carey Foster Bridge, metre bridge, Wien bridge etc.

The Principle of Wheatstone Bridge

The circuit of the Wheatstone bridge consists of four resistors P, Q, R, S and they are arranged as the quadrilateral ABCD. Points A and B are connected to battery E through K₁. The B and D points are connected to the galvanometer G via the key K₂. The resistances are selected such that the galvanometer needle doesn’t deflect or current Ig through it is 0. This is essentially called the balanced condition of a bridge or null condition. For this condition

\[ \frac{P}{Q} = \frac{R}{S} \]

Instead of S, the unknown resistor is connected and resistor R can be varied. Resistors P and Q are also known as ratio arms. You keep a fixed value for the ratio \[ \frac{P}{Q} \] and then R is adjusted to the value where the null condition is met. Unknown resistance is given by the \[S = \frac{Q}{P} R \] .

Derivation of Wheatstone Bridge

The current through the galvanometer, at the balanced condition of the bridge, is zero, which is  IG = 0. Current through arms AB and BC is denoted by I1 Current through arms of AD as well as DC is I2 As per the circuital law of Kirchhoff, voltage drop through the closed-loop is 0. The application of this law in the ABDA loop results in the dropping of the sum of voltage through the individual arms of the loop which is 0.

\[ I_{1} P - I_{2} R = 0 \]

\[ \frac{I_{1}}{I_{2}} = \frac{R}{P} \]

Applying Kirchhoff’s law to the loop CBDC

\[ I_{1} Q - I_{2} S = 0 \]

\[ \frac{I_{1}}{I_{2}} = \frac{S}{Q}\]

Comparing conditions,

\[ \frac{R}{P} = \frac{S}{Q} \]

\[ \frac{P}{Q} = \frac{R}{S} \]

This is the formula for Wheatstone bridge.

Application of Wheatstone Bridge

• Resistance measurement via direct application of Ohm’s law can’t be done precisely. In this setup, current and voltage through the unknown resistor must be measured using an ammeter and voltmeter respectively. The ideal ammeter must have zero resistance plus the ideal voltmeter must have infinite resistance. However infinite or zero resistance is impossible and therefore the circuit is unable to give precise measurements. Here the whitestone bridge circuit can be deployed for accurate measurements in such scenarios. For experimental purposes, two adaptations of the Wheatstone bridge circuit are the Metre bridge and Post office box. 

• Resistance of some of the materials, such as semiconductors, tends to vary with the temperature. The variations are large in comparison to the ordinary resistors. These are known as thermistors. A slight change in temperatures can be measured using thermistors for the Wheatstone bridge setup.

• Changes in the intensity of light are measured via the replacement of an unknown resistor, in the Wheatstone bridge circuit, with a photoresistor. The resistance of the photoresistor is the function of the incident light.

• Wheatstone bridge can also be used for measuring strain and pressure. 

Limitations of Wheatstone Bridge

• Wheatstone bridge is a very sensitive device. The measurements may not be precise in an off-balance condition.

• Wheatstone bridge is generally used for measuring resistances ranging from a few ohms to a few kilo-ohms. 

• The sensitivity of the circuit reduces if the four resistances are not comparable.

Solved Examples

1. The ratio arms of a Wheatstone bridge has resistances equal to 100 \[ \Omega \] and 10\[ \Omega \]. An unknown resistor is connected to the fourth arm. What should be the value of the unknown resistance if the third arm has a resistance of 153 \[ \Omega \] in a balanced condition?

Solution: Resistance of the first arm P=100 \[ \Omega \]

Resistance of the second arm Q=10\[ \Omega \]

Resistance of the third arm R=153\[ \Omega \]

If the unknown resistance is X, the ratio of resistances in the balanced condition If the unknown resistance is X, the ratio of resistances in the balanced condition,

\[\frac{R}{X} = \frac{P}{Q} \]

\[ X = \frac{Q}{P} R\]

Substituting the values,

\[ X = \frac{10}{100} 153 \Omega\]

\[ X = 15.3 \Omega \]

The unknown resistance is 15.3 \[ \Omega \]

2. The four resistances of a Wheatstone bridge are 100\[ \Omega \], 10\[ \Omega \], 300\[ \Omega \], and 30\[ \Omega \]. If the bridge is connected to a 1.5 V battery, what are the currents through individual resistors?

Solution:  Resistance of the first arm P=100\[ \Omega \]

Resistance of the second arm Q=10\[ \Omega \]

Resistance of the third arm R=300\[ \Omega \]

Resistance of the fourth arm S=30\[ \Omega \]

The image will be uploaded soon

Points A and C are connected to the battery such that, the potential difference is VAC =1.5V.

The ratios of the arms,

\[ \frac{P}{Q} = \frac{100}{10} = 10 \]

\[ \frac{R}{S} = \frac{300}{30} = 10\]

Therefore, the null condition is satisfied,

\[ \frac{P}{Q} = \frac{R}{S} \].

The current through the galvanometer is zero. The total resistance along the path ABC is R1=P+Q since these two resistances are connected in series. Similarly, total resistance along the path ADC is

\[ R_{2} = R + S \]. The resistance R1 and R2 are connected in a  parallel combination between points A and C. Therefore,

Current through P= current through \[Q= I_{1}\]

\[I_{1} = \frac{V_{AB}}{R_{1}} \]

\[= \frac{V_{AB}}{P+Q} \]

\[= \frac{1.5V}{(100 + 10)\Omega} \]

= 0.0136A

Current through R = current through \[S = I_{2}\] where

\[I_{2} = \frac{V_{AB}}{R_{2}} \]

\[\frac{V_{AB}}{R+S}\]

\[\frac{1.5 V}{(300+30)}\Omega\]

= 0.0045A

The current through the 100\[ \Omega \] and 10\[ \Omega \] resistors is 0.0136 A whereas the current through the 300\[ \Omega \] and 30\[ \Omega \] resistors is 0.0045 A.

Did You know?

• The Wheatstone bridge circuit was initially invented by Samuel Hunter Christie and later improved by Charles Wheatstone.

• Various adaptations of the Wheatstone bridge can be used to measure impedance, inductance, and capacitance in AC circuits.

• Maxwell improved the circuit to use for AC circuits, which is known as the Maxwell bridge.