How does a bridge circuit work?

The standard bridge circuit is known as a Wheatstone Bridge.
  • Wheatstone bridge is used to determine an unknown resistance by balancing the two legs of the bridge circuit, of which one leg includes the unknown resistance component.
  • Wheatstone bridge works on the null deflection principle, where the ratio of resistances is equal and no current flows through it.
  • Under normal conditions, the bridge remains unbalanced and the current flows through the galvanometer.
  • In a balanced condition, no current flows through the galvanometer by adjusting the values of the known and variable resistance.
  • When the current through a galvanometer is zero, the following condition exists I1P = I2R.

Wheatstone Bridge

To measure unknown resistance, a bridge circuit consists of a resistor with an unknown value and three resistors of known value. Either of the three known resistors is adjusted and then replaced until the bridge is balanced. When the balance has been reached, the unknown resistance can be determined from the ratio of the known resistances.


Related Questions

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

Read More:

CBSE CLASS XII Related Questions

  • 1.
    If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.


      • 2.
        The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

          • \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
          • \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
          • \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
          • Zero

        • 3.
          Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.


            • 4.
              Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

                • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
                • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
                • Assertion (A) is true, but Reason (R) is false.
                • Both Assertion (A) and Reason (R) are false.

              • 5.
                Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.


                  • 6.
                    Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

                      • \((P_1+P_2)\)
                      • \(\dfrac{P_1+P_2}{2}\)
                      • \(\dfrac{P_1P_2}{2(P_1+P_2)}\)
                      • \(\dfrac{P_1P_2}{4(P_1+P_2)}\)
                    CBSE CLASS XII Previous Year Papers

                    Comments


                    No Comments To Show