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

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                    CBSE CLASS XII Previous Year Papers

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