Wheatstone Bridge

What is Wheatstone Bridge?

Wheatstone bridge is a setup to measure an unknown resistance. A Wheatstone bridge has four arms (resistors) and the ratio of two of the resistors is kept at a fixed value. The other two arms are balanced, one of which is the unknown resistor whereas the resistance of the other arm can be varied. The unknown resistance is computed using the balancing or null condition. The Wheatstone bridge circuit gives a very precise measurement of resistance. Various adaptations of the Wheatstone bridge are used for AC circuits. Some instruments based on the Wheatstone bridge principle are meter bridge, Carey Foster bridge, Wien bridge, etc.    

Wheatstone Bridge Principle

The Wheatstone bridge circuit is shown in the above figure. Four resistors P, Q, S, R  are arranged as a quadrilateral ABCD. The points A and B are connected to a battery E through the key \[K_{1}\]. The points B and D are connected to a galvanometer G through the key \[K_{2}\].

The resistances are so chosen that the galvanometer needle does not deflect or the current \[I_{G}\]. through it is zero. This is called the null condition or the balanced condition of the bridge. At this condition,

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

The unknown resistor is connected instead of S and the resistor R can be varied. The resistors P and Q are sometimes referred to as the ratio arms. The ratio P/Q is kept fixed and R is adjusted to a value such that the null condition is met. The unknown resistance is given by,

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

Wheatstone bridge derivation

At the balanced condition of the bridge, current through the galvanometer is zero i.e. \[I_{G}\] = 0. Current through the arms AB and BC is \[I_{1}\]. Current through the arms AD and DC is \[I_{2}\]. According to Kirchhoff’s circuital law, the voltage drop across a closed loop is zero.

Applying Kirchhoff’s law in the loop ABDA, the sum of voltage drops across the individual arms of the loop is zero i.e.

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

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

Applying Kirchhoff’s law in the loop CBDC,

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

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

Comparing the conditions,

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

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

This is the Wheatstone bridge formula. 

Application of Wheatstone Bridge

  • The measurement of resistance through direct application of Ohm’s law can not be done precisely. In such a setup, the current and voltage across the unknown resistor should be measured using an ammeter and a voltmeter respectively. An ideal ammeter should have zero resistance and an ideal voltmeter should have infinite resistance but practically zero or infinite resistance is impossible. Therefore, this circuit cannot give precise measurements. Wheatstone bridge circuit can be employed for very precise measurements in such cases. Two adaptations of the Wheatstone bridge circuit for experimental purposes are,

  • Post office box

  • Meter bridge

  • The resistance of some materials (e.g. semiconductors) varies with temperature. The variations are quite large compared to ordinary resistors. These are called thermistors.  Slight changes of temperature can be measured using thermistors in the Wheatstone bridge setup.

  • Changes in light intensity can be measured by replacing the unknown resistor, in a Wheatstone bridge circuit, with a photoresistor. The resistance of a photoresistor is a function of incident light. 

  • Wheatstone bridge can also be used to measure 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,

 \[\frac{R}{X}\] = \[\frac{P}{Q}\]

X =  \[\frac{Q}{P}\]R

Substituting the values,

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

X=15.3\[\Omega\]

The unknown resistance is 15.3\[\Omega\]. 

  1. 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\[\Omega\]

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The points A and C are connected to the battery such that, the potential difference is \[V_{AC}\] =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 \[R_{1}\]=P+Q, since these two resistances are connected in series. Similarly, total resistance along the path ADC is \[R_{2}\]=R+S

The resistances \[R_{1}\]and \[R_{2}\] are connected in a  parallel combination between the points A and C. Therefore,

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

 \[I_{1}\] = \[\frac{V_{AB}}{R_{1}}\]

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

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

    =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 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 Maxwell bridge.

FAQ (Frequently Asked Questions)

1. What is Wheatstone bridge used for?

Wheatstone bridge is used to measure resistances ranging from few ohms to few kilo-ohms. The measurements are very precise since the apparatus is very sensitive. Some arrangements, based on the same principle, are

  • Carey Foster bridge

  • Post office bridge

  • Meter bridge

  • Kelvin bridge

  • Maxwell bridge

  • Wien bridge

2. Why are Wheatstone bridge measurements accurate?

The common setups lack precision because practical ammeters and voltmeters do not have zero and infinite resistances respectively. The principle of Wheatstone bridge is based on the null method (the arrangement is such that the current through the galvanometer is zero) that does not depend on the resistance of the galvanometer. This makes the measurements very precise.