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?

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CBSE CLASS XII Related Questions

  • 1.
    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:

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

    • 2.

      A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

        • 80 Wb
        • 16 Wb
        • \(8 \times 10^{-2}\) Wb
        • \(8 \times 10^{-3}\) Wb

      • 3.
        A wire of length \(L\) is bent round into (i) a square coil having \(N\) turns and (ii) a circular coil having \(N\) turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.


          • 4.
            Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.


              • 5.

                A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:

                (I) the formation of maxima and minima in the diffraction pattern, and

                (II) why the maxima go on becoming weaker and weaker with its increasing number $n$.


                  • 6.

                    A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:

                    (I) the magnetic dipole moment of the coil, and

                    (II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

                      CBSE CLASS XII Previous Year Papers

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