A galvanometer having resistance 100 ohms shows full scale deflection with current 10 mA. Find value of shunt to convert it into an ammeter of 10 ampere range.

A galvanometer can be defined as a device used for the detection of small electric current or to measure its magnitude. In general, the current and its intensity is shown by the movement of a magnetic needle or a coil in a magnetic field. When current IG flows via the galvanometer, the current through the shunt resistance can be expressed by IS = I – IG. The voltages across the galvanometer and shunt resistance are equivalent to one another due to the parallel nature of their connection.

Therefore, RG.IG = (I – IG).Rs

Here,

  • RG = Resistance of galvanometer
  • G = Galvanometer coil
  • I = Total current via circuit
  • IG = Total current via galvanometer that corresponds to full-scale reading
  • Rs = Value of shunt resistance

Step-1: Solution

As per the given question, the following can be determined:

I = 10A
Ig = 10 mA = 0.010 A
Rg = 100 Ω

Thus, with the given condition, we can obtain:

I = Ig + Is

⇒ (I − Ig)Rs = IgRg

By replacing the known values, we can get:

(10 − 0.010)Rs = 0.010 × 100

⇒ Rs = 0.1

Also Read: NCERT Solutions for Class 6 to 12


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CBSE CLASS XII Related Questions

  • 1.
    Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

      • The total charge of the two spheres is conserved.
      • Both spheres attain the same potential.
      • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
      • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

    • 2.
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        • 3.
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            • 4.
              With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.


                • 5.
                  Read the following paragraph and answer the questions that follow.
                  A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


                    • 6.
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                        CBSE CLASS XII Previous Year Papers

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