Wheatstone’s Bridge, Meter Bridge and Potentiometer

Finding Unknown Resistance using Wheatstone Bridge

Wheatstone’s Bridge

  • Developed by Sir Charles Wheatstone in 1843
  • Use of a bridge circuit to measure unknown resistance
  • Consists of 4 resistors arranged in a diamond shape

Wheatstone’s Bridge Circuit

``

R1 R2
V R R IR
S 3 4 
     |        |
    R3       Rx

``

Working Principle of Wheatstone’s Bridge

  • Condition for balanced bridge:
    (R1/R2) = (R3/Rx)
  • When the bridge is balanced:
    • No current flows through the galvanometer
    • Ratio of resistances is equal

Applications of Wheatstone’s Bridge

  • Determining unknown resistance
  • Measuring small variations in resistance
  • Null detection in sensitive laboratory instruments

Meter Bridge

  • Also known as slide wire bridge or potentiometer
  • Consists of a long uniform wire stretched over a wooden board
  • Used to find the unknown resistance and verify Ohm’s law

Circuit Diagram of Meter Bridge

P - Q | | R1 X | | - J | | R2 Y | | - K | | G M N H

Working of Meter Bridge

  • Jockey is moved along the wire until the null point is obtained
  • No current flows through the galvanometer when balanced
  • Ratio of resistances is equal: (R1/X) = (R2/Y)

Potentiometer

  • Device used to measure emf and potential difference
  • Consists of a long uniform wire
  • Comparing unknown emf with a known standard cell emf

Circuit Diagram of Potentiometer

`` A BC D | | | E X-Y F | | | G | |

     |             |
    J              K
     |             |
   R1               Rx
     |             |

``

Properties of a Potentiometer

  • Measures potential difference accurately
  • No current flows in the potentiometer wire
  • Can measure emf of a cell directly

Balancing Length in Potentiometer

  • Balancing length is the position on the wire where no current flows through the galvanometer
  • At a balanced point, potential drop across a length is equal to the emf of the cell being measured

Formula for Balancing Length

  • Balancing length (L) is given by: L = (E × l) / V where, E is the emf of the cell, l is the length of the potentiometer wire, V is the potential gradient of the potentiometer wire

Wheatstone Bridge Vs. Potentiometer

Wheatstone Bridge:

  • Measures unknown resistance
  • Comparison of resistances
  • Needs a known resistance Potentiometer:
  • Measures potential difference or emf
  • Does not require a known resistance
  • No power loss in the circuit

Advantages of Wheatstone’s Bridge

  • Highly accurate method of measuring resistance
  • Can be used for both high and low resistances
  • Easy to use and portable

Advantages of Meter Bridge

  • High sensitivity and accuracy in measuring unknown resistances
  • Directly verifies Ohm’s law
  • Can be used for both DC and AC circuits

Advantages of Potentiometer

  • Highly accurate in measuring potential difference and emf
  • No current flow through the wire, reducing errors
  • Can be used with different cells or batteries

Limitations of Wheatstone’s Bridge

  • Requires a known resistance for comparison
  • Works only if the galvanometer has high sensitivity

Limitations of Meter Bridge

  • Difficult to obtain exact null point due to thermal effects
  • Not suitable for low resistance measurements

Limitations of Potentiometer

  • Requires a long wire, limiting portability
  • Measurement errors can occur due to wire resistivity and faults in connections

Slide 21:

  • Example: Determine the value of the unknown resistance Rx in a Wheatstone bridge circuit having R1 = 100 ohms, R2 = 200 ohms, and R3 = 150 ohms.
  • Equation: (R1/R2) = (R3/Rx)
  • Substituting the given values, we have (100/200) = (150/Rx)
  • Solving for Rx, we find Rx = 300 ohms

Slide 22:

  • Example: A meter bridge has a wire of length 100 cm. The resistances R1 and R2 are 10 ohms and 20 ohms respectively. Find the balancing length when the unknown resistance X is 5 ohms.
  • Equation: (R1/X) = (R2/Y), where Y is the length from one end of the wire to the null point
  • Substituting the given values, we have (10/5) = (20/Y)
  • Solving for Y, we find Y = 10 cm, which is the balancing length

Slide 23:

  • Example: A potentiometer has a wire of length 200 cm and a potential gradient of 0.1 V/cm. If the emf of a cell being measured is 1.5 V, find the balancing length.
  • Equation: L = (E × l) / V, where L is the balancing length, E is the emf, l is the length of the wire, and V is the potential gradient
  • Substituting the given values, we have L = (1.5 × 200) / 0.1 = 300 cm

Slide 24:

  • Example: A student measures the balancing length of a potentiometer wire to be 40 cm. If the potential gradient is 0.2 V/cm, determine the emf of the cell being measured.
  • Equation: E = (L × V) / l, where E is the emf, L is the balancing length, V is the potential gradient, and l is the length of the wire
  • Substituting the given values, we have E = (40 × 0.2) / 100 = 0.08 V

Slide 25:

  • Example: In a Wheatstone bridge circuit, if R1 = 5 ohms and R2 = 10 ohms, what should be the value of R3 to balance the bridge when the unknown resistance Rx is 15 ohms?
  • Equation: (R1/R2) = (R3/Rx)
  • Substituting the given values, we have (5/10) = (R3/15)
  • Solving for R3, we find R3 = 7.5 ohms

Slide 26:

  • Example: A meter bridge has a wire of length 50 cm. The resistances R1 and R2 are 5 ohms and 10 ohms respectively. Find the balancing length when the unknown resistance X is 8 ohms.
  • Equation: (R1/X) = (R2/Y), where Y is the length from one end of the wire to the null point
  • Substituting the given values, we have (5/8) = (10/Y)
  • Solving for Y, we find Y = 16.7 cm, which is the balancing length

Slide 27:

  • Example: A potentiometer has a wire of length 150 cm and a potential gradient of 0.05 V/cm. If the emf of a cell being measured is 2 V, find the balancing length.
  • Equation: L = (E × l) / V, where L is the balancing length, E is the emf, l is the length of the wire, and V is the potential gradient
  • Substituting the given values, we have L = (2 × 150) / 0.05 = 600 cm

Slide 28:

  • Example: A student measures the balancing length of a potentiometer wire to be 30 cm. If the potential gradient is 0.3 V/cm, determine the emf of the cell being measured.
  • Equation: E = (L × V) / l, where E is the emf, L is the balancing length, V is the potential gradient, and l is the length of the wire
  • Substituting the given values, we have E = (30 × 0.3) / 100 = 0.09 V

Slide 29:

  • Applications of Wheatstone Bridge:
    • Measurement of unknown resistance
    • Determination of unknown resistivity of a material
    • Calibration of resistance measuring instruments

Slide 30:

  • Applications of Meter Bridge:
    • Verification of Ohm’s law
    • Measurement of small resistances
    • Measurement of unknown resistance by null deflection technique