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What is the use of the meter bridge experiment?

Use of meter bridge experiment:

The meter bridge experiment is used to test the laws of resistance and figure out unknown resistances.

  • Meter Bridge wire is made of copper-nickel alloy (constantan), alloy manganin, and chromium, nickel, and iron alloys (nichrome) as they are resistant to corrosion.
  • They have cold temperatures and a high melting point of around 1400 degrees Celsius.
  • A meter bridge is an electrical device that is used to calculate the resistance of a conductor.

It can also be used to determine the series and parallel equivalent of resistances.

Related Questions

  1. Can you find very high resistance accurately with the help of a Metre bridge?
  2. Why Carey Foster Bridge Is So Sensitive?
  3. Meter Bridge or Slide Wire Bridge is a practical form of?
  4. Meter bridge works on the principle of?
  5. Why should we get the null point in the middle of the Metre bridge wire?
  6. How does a bridge circuit work?
  7. How do you solve an unbalanced bridge?
  8. Explain the construction of meter bridge.
  9. Why is Wheatstone's bridge more accurate?
  10. What Is Null Voltage?

Read More:

Wheatstone Bridge – Important Topics
Unit of Specific Resistance Resistivity Temperature Dependence Internal resistance formula
Current density Power Of Alternating Current Resistor Colour Codes
Voltage Difference between kva and kw Electrical Formula
Ohm’s law Relation between kva and kw Combination of Resistors
Carbon Resistor Difference between AC and DC Potentiometer
Electromotive Force Ohm’s Law and its Limitations Derivation of Drift Velocity
Electric Dipole Cells, emf, Internal Resistance Permittivity And Permeability
Types of Current Differences Between Acceleration and Velocity Power transformers
Wheatstone Bridge Relation between resistance and length Static Electricity
Resistivity formula Difference between RTD and thermocouple Current density formula
Ampere Difference between air conditioning and refrigeration Types of Resistors
Resistance Formula Difference Between Primary Cell and Secondary Cell Current Electricity
Drift Velocity Relation between power and resistance Cells in Series and in Parallel
Conductivity Difference between emf and voltage Types of Circuits
Unit of Current Difference between earth and neutral Uses of Resistor

CBSE CLASS XII Related Questions

  • 1.

    (I) The real image of an object placed between $f$ and $2f$ from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? Explain.

    (II) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

        View Solution

      • 2.
        In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

          • \(10\ \text{N/C}\) along \(+x\)
          • \(10\ \text{N/C}\) along \(-x\)
          • \(50\ \text{N/C}\) along \(+x\)
          • \(50\ \text{N/C}\) along \(-x\)
          View Solution

        • 3.

          A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

            • 5 V
            • 3.5 V
            • 2.5 V
            • Zero V
            View Solution

          • 4.

            A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

              • the induced current will flow in the anticlockwise direction.
              • the induced current will flow in the clockwise direction.
              • no induced current will flow in the solenoid.
              • the magnet will fall with a constant velocity.
              View Solution

            • 5.
              Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

                • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
                • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
                • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
                • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)
                View Solution

              • 6.
                A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

                    View Solution
                  View All

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