Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is? (A). 250 Ω (B). 40 Ω (C). 500 Ω (D). 25 Ω

 Current and voltage sensitivity is given by

IS = nBA/c……………………(1)

VS = nBA/cR…………………….(2)

We can find the current sensitivity of the galvanometer.

The current sensitivity of the galvanometer is given by, I= nBA/c……………………(3)


  • n is the number of turns in the coil of galvanometer,
  • B is the magnetic field around the coil
  • A is the area of the coil
  • c is the restoring torque per unit twist

It is given that current sensitivity is 5 div/mA.

We can put this value in equation (3).

So, we can write that, IS=nBA/c=5 in (2) gives the voltage sensitivity of the galvanometer,

VS = nBA/cR

The voltage sensitivity of the galvanometer is 20div/V.

Now, we will look for a relation between the voltage sensitivity and current sensitivity of the galvanometer.

Combining Equation (1) and (2) will give us, VS= nBA/cR = 20 So, the resistance of the galvanometer is ,

VS = IS /R

⇒ 20 = 5/10−3/R

⇒ R = 250

So, the resistance of the galvanometer is- 250 Ω

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