Wheatstone Bridge: Derivation, Formula & Applications

Collegedunia Team logo

Collegedunia Team

Content Curator

Wheatstone bridge is an electric circuit that is used to measure an unknown resistance by balancing two legs of the circuit. The circuit is used to provide accurate measurements. Wheatstone bridge consists of four arms (R1, R2, R3, and R4) of which two arms have known resistances.

  • The other two arms consist of an unknown resistance and a variable resistance. 
  • Wheatstone Bridge gives an extremely accurate measurement as compared to the simple voltage divider.
  • The principle of working of Wheatstone bridge is similar to a potentiometer.
  • The Wheatstone bridge works on the null deflection principle and is an application of Kirchhoff’s law.

Key Terms: Wheatstone bridge, Resistance, Circulation of current, Resistors, Balanced bridge circuits, Circuits, Electrical circuit, Voltage divider


What is a Wheatstone Bridge?

[Click Here for Sample Questions]

Wheatstone bridge is used to calculate the unknown resistance precisely. It consists of four resistors that are connected in a diamond shape with the DC supply source connected across the top and bottom points (C and D in the circuit) of the diamond and the output is taken across the other two ends (A and B in the circuit).

  • The Wheatstone Bridge was invented in 1833 by Samuel Hunter Christie. 
  • Later, it was improved by Sir Charles Wheatstone in 1843.
  • The Wheatstone bridge is composed of four resistances P, Q, R, and X.
  • Two legs have fixed resistance P and Q.
  • Of the remaining two legs, one includes the component of the unknown resistance x.
  • One variable resistance, R is also connected on the fourth leg of the Wheatstone bridge.

Wheatstone bridge

Wheatstone Bridge

  • In the above Wheatstone bridge diagram, E is the battery, G is the galvanometer, and K is the Tapping key which helps in completing the circuit so that current may flow through the circuit.

Discover about the Chapter video:

Current Electricity Detailed Video Explanation:

Also Read:


Working Principle of Wheatstone Bridge

[Click Here for Previous Year Questions]

The Wheatstone bridge works on the principle of null deflection i.e. the ratio of resistances is equal and no current flows through the electric circuit.

  • Under normal conditions, the Wheatstone bridge is in an unbalanced condition where current flows through the galvanometer.
  • The bridge is in a balanced condition when no current flows through the galvanometer by adjusting the variable and the known resistance. 

Wheatstone Bridge Formula

[Click Here for Sample Questions]

The formula used for the Wheatstone bridge is:

X = \(Q \times {R \over P}\)
 

Where,

  • X is the unknown resistance
  • Q is the standard arm of the bridge
  • R and P is the ratio of the arm of the bridge

Wheatstone Bridge Derivation

[Click Here for Previous Year Questions]

Wheatstone Bridge Formula can be derived by – 

  • Current I which originates from the battery E passes through the circuit.
  • When current I reach point A, it gets divided into two parts – current I1 and current I2.
  • Current I1 passes through arm AB and current I2 passes through arm AD.
  • When current I1 passes through point B it is further divided into two parts I3 and I4.
  • According to Kirchhoff’s law, we know that. 

I4 = I1 - I3

Diagram 2

Working of Wheatstone Bridge

  • I2 current passes through the arm AD and gets combined with the current I3, together they make current I5 which moves through the DC arm.

I5 = I2 + I3

  • At point C both the current that is current I4 and I5 combine and as a result of which we get current I. As we know.
I4 = I1 - I3 and I5 = I2 + I3
  • Now when we combine both the values of I4 and I5 we get

I4 + I5 = I1 - I3 + I2 + I3

  • Now, when we eliminate the value of current I3 from the above equation we get I4 + I5 = I1 +I2 and we know that I1 + I2 = I(According to Kirchhoff law )
  • Throughout the circulation of current in the circuit, we must keep changing the value of R till there is no deflection in the galvanometer (G).
  • When there is no deflection in the galvanometer then the value of I3 becomes zero(0).
  • In this condition in which the value of I3 becomes zero or there is no deflection in the galvanometer, this particular condition is referred to as the balanced condition of the Wheatstone bridge

VA – VB = VAB (according to ohm's law V = IR)

Now,

VAB = VA – VB = I1P --- (Equation 1)

VBC = VB – VC = I1Q---(Equation 2)

VAD = VA – VD = I2R---(Equation 3)

VDC = VD – VC = I2X --- (Equation 4)

  • Since we know that the potential difference between B and D will be zero VB –V= 0

VBD = G*I3 ( according to ohm's law )

  • And we know that I3 = 0 therefore the potential difference between VBD becomes zero that is VBD = 0. So VB = VD.
  • On comparing the Equation 1 & 3, we get I1P = I2R --- (Equation 5)
  • On comparing the Equation 2 & 4, we get I1Q = I2X --- ( EQUATION 6)

On dividing equation 6 by equation 5 we get the following result.

\({Q \over P} = {X \over R}\)

Now from the above equation, we get X = Q*R/P


Applications Of Wheatstone Bridge

[Click Here for Sample Questions]

Wheatstone bridges have applications in many different electronic applications such as– 

  • The measurement of resistance via direct application of Ohm’s law can’t be determined precisely with the Wheatstone bridge. In this setup, current and voltage through the unknown resistor need to be measured using an ammeter and voltmeter respectively. The ideal ammeter has zero resistance whereas the ideal voltmeter has infinite resistance. However infinite or zero resistance is not possible and therefore precise measurements cannot be found. The Wheatstone bridge circuit is deployed here for accurate measurements. 
  • Changes in the intensity of light are measured by replacing an unknown resistor, in the Wheatstone bridge, with a photoresistor. The resistance of the photoresistor is the function of the incident light.
  • The resistance of some materials like semiconductors varies with the temperature. In comparison to the ordinary resistors, these variations are large and hence they are known as thermistors. A slight change in temperatures can be measured using thermistors through the Wheatstone bridge circuit.
  • A wheatstone bridge can also be used to measure strain and pressure. 

Read more: NCERT Solutions Current Electricity


Advantages of Wheatstone Bridge

[Click Here for Previous Year Questions]

The advantage of balanced bridge circuits or Wheatstone bridges is as follows:

  • One of the main advantages of the Balanced bridge circuits or Wheatstone bridge is that it can be easily interfaced into various combinations.
  • The Wheatstone bridge is traditionally called ohmmeter as the results are measured in terms of resistance and also are accurate and precise.
  • One can easily measure the minute changes in the bridge, even in ohms' if needed.
  • It becomes very easy to find out the value of resistance that is not known beforehand as the value of the other three resistance is known to us.

Limitations of Wheatstone Bridge

[Click Here for Sample Questions]

Some of the limitations of Wheatstone Bridge are – 

  • It is a very sensitive device. In an off-balance condition, the measurements may not be precise.
  • Wheatstone bridge is generally used to measure resistances ranging from a few ohms to a few kilo-ohms. 
  • The sensitivity of the circuit reduces if the four resistances of the circuit are not comparable.

Read More: Current Electricity Important Questions


Variations of Wheatstone Bridge

[Click Here for Previous Year Questions]

Various adaptations of the Wheatstone bridge are used to measure inductance, impedance, and capacitance in AC circuits. These adaptations include- 

  • Carey Foster bridge 
  • Post office bridge
  • Meter Bridge
  • Kelvin bridge 
  • Maxwell bridge
  • Wien bridge

Read more:


Previous Year Questions 


Things to Remember

  • The Wheatstone Bridge is one of the simplest bridge circuits, which can be used to measure resistance very precisely.
  • The wheatstone bridge was invented by Samuel Hunter Christie in 1833, which Sir Charles Wheatstone later popularised in 1843.
  • It comprises two known resistors, one unknown resistor, and one variable resistor connected in the form of a bridge. 
  • The Wheatstone bridge works on the principle of null deflection, meaning the ratio of their resistances is equal and no current flows through the circuit. 
  • The formula used for the Wheatstone bridge is, R = PS/Q, 

    Where R is the unknown resistance, S is the standard arm of the bridge, and P and Q are the ratios of the arm of bridges.


Sample Questions

Ques. When can a Wheatstone Bridge be considered unbalanced? (1 mark)

Ans. Wheatstone Bridge can be considered unbalanced when the current flows through the galvanometer bridge under normal conditions.

Ques. A Wheatstone bridge consists of four resistances 200Ω, 20Ω, 400Ω, and 40Ω. If the bridge is connected to a 1.5 V battery, then find the currents through individual resistors. (5 marks)

Ans. Given:

First arm resistance P=200Ω

Second arm resistance Q=20Ω

Third arm resistance R=400Ω

Fourth arm resistance S=40Ω

The potential difference is VAC =1.5V as points A and C are connected to the battery.

Therefore, the ratios of the arms,

P/Q=200/20 = 10

R/S= 400/40 = 10

Hence it satisfies the null condition, P/Q = R/S

∴ P = VAB/(P+Q)

That is, 1.5/ (200+20) = 0.0681A

Again, current in the resistance R = current in the resistance VAB/(R+S)

That is, 1.5/ (400+40) = 0.0340A

Ques. 300 Ω and 30Ω are the resistances of the ratio arms of a Wheatstone bridge. The fourth arm is connected to an unknown resistor. Calculate the value of the unknown resistance if the third arm has a resistance of 250Ω in a balanced condition. (2 marks)

Ans. Given:

First arm resistance P=300Ω

Second arm resistance Q=30Ω

Third arm resistance R=250Ω

Let us assume, the unknown resistance is X, and the ratio of resistances in the balanced condition will be, R/X=P/Q

X=(Q/P)

By substituting the values we get,

X= (30/300)250Ω

X= 25Ω

Ques. Write is the principle of the Wheatstone Bridge. (1 mark)

Ans. The working of a Wheatstone bridge basically depends on the principle of null deflection, which means that the ratio of their resistances is equal, and no current flows through the circuit.

Ques. Which among the following is a false statement? a) A galvanometer is used as the null detector in a Wheatstone bridge; b) A galvanometer is an ammeter with low resistance in series; c) Wheatstone bridge is susceptible to high dc current; d) Due to the errors introduced in contact resistance, a Wheatstone bridge cannot be used for accurate measurement. (2 marks)

Ans. Correct Option: (c)

Explanation: A Wheatstone bridge is not susceptible to high dc current, but if not balanced, it can give inaccurate readings.

Ques. What are the applications of a Wheatstone bridge? (4 marks)

Ans. The applications of a Wheatstone bridge are –

  • Measurement of low resistance by means of direct application of Ohm’s law can not be done accurately. In such cases, Wheatstone bridge circuits can be installed for accurate measurements.
  • Some materials such as semiconductors vary in resistance in relation to the temperature, and as compared to ordinary resistors, these variations are huge. These materials are recognized as thermistors. A minor change in temperatures can be measured by using thermistors for the Wheatstone bridge arrangement.
  • A Wheatstone bridge circuit is used to measure the changes in the intensity of light using the replacement of an unknown resistor with a photoresistor. The photoresistor functions on the basis of incident light.
  • A Wheatstone bridge helps in measuring strain and pressure.

Ques. What are the limitations of the Wheatstone Bridge? (4 marks)

Ans. The limitations of the Wheatstone Bridge are:

  • It is a very sensitive device which is why it is possible that the measurements using a Wheatstone bridge may not be precise in an off-balance condition.
  • Usually, the Wheatstone bridge measures resistances that are in limited ranges from ohms to kilo-ohms which means that for high resistance measurements, the galvanometer is insensitive to imbalance.
  • The other limitation is that the value of resistance changes; therefore, extreme current may even cause a permanent change in the value of resistance due to the heating effect of the current through the resistance.
  • The sensitivity of the circuit reduces if the four resistances are not comparable.

Ques. When electrons drift in metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? (Delhi 2012)
electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction

Ans. No, this is because only the drift velocities of the electrons are superposed over their random thermal velocities. The solid line depicts the random path followed by a free electron in the absence of an external field. The electron proceeds from A to B, making six collisions on its path. And the dotted curve shows the way in which the random motion of the same electron gets modified when an electric field is applied.

Ques. Give reasons for Why the emf of a cell is always greater than its terminal voltage? (Delhi 2013)

Ans. When no current is drawn Emf is the p.d. When current is drawn, there will be a potential drop across the internal resistance of the cell. So, the terminal voltage will be less than the emf.

Ques. I – V graph for a metallic wire at two different temperatures, T1 and T2. Which of the two temperatures is lower and why? (All India 2015)

Ans. The temperature T1 is lower. Larger the slope of the V-I graph, the smaller the resistance. This is because the resistance of a metal increases with the increase of temperature, resistance at temperature T1 is lower.

Ques. In the electric network, use Kirchhoff's rules to calculate the power consumed by the resistance R = 8 Ω. (Comptt. Delhi 2012)

Ans. In loop BCDA,
I1 × 4 + (I1 + I2) × 8 = 12
4I1 + 8I1 + 8I2 = 12
12I1 + 8I2 = 12
∴ 3I1 + 2I2 = 3 … (Dividing by 4) … (i)
In loop ADFE,
(I1+ I2) × 8 = 8 => 8I1, + 8I2 = 8
∴ I1 + I2 = 1 …(Dividing by 8)
Solving equations (i) and (ii), we get
I1 = 1A and I2 = 0A
∴ Power consumed in 80 resistance (R) = I2R
= (I1 + I2)2 × R
= (1 + 0)2 × 8 = 8 watt

Ques. Which among the following ratios defines the sensitivity of a galvanometer? a) Deflection in the galvanometer to the unit change in unknown resistance b) Unit change in unknown resistance to thrice the deflection in the galvanometer c) Half the unit change in unknown resistance to the deflection in the galvanometer d) Deflection in the galvanometer to twice the unit change in unknown resistance. (3 marks)

Ans. The correct option is (a)

Explanation: The sensitivity of a galvanometer can be defined as the ratio of deflection in the galvanometer to the unit change in unknown resistance. The sensitivity of a galvanometer can be increased by using a strong magnet or increasing the area of the coil by increasing the number of turns.

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Also Read:

CBSE CLASS XII Related Questions

1.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field ?
  1. inside the sphere
  2. just outside the sphere
  3. at a point 18 cm from the centre of the sphere?

      2.
      A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is

        • 0
        • 1 m

        • 2 m

        • 3 m

        3.

        A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.

        1. What is the rms value of the conduction current?
        2. Is the conduction current equal to the displacement current?
        3. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
        A parallel plate capacitor made of circular plates

            4.

            In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

                5.

                Three capacitors each of capacitance 9 pF are connected in series. 

                (a) What is the total capacitance of the combination? 

                (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

                    6.

                    A tank is filled with water to a height of 12.5cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

                        CBSE CLASS XII Previous Year Papers

                        Comments



                        No Comments To Show