Question

A galvanometer has current range of 15 ma and voltage range 750 mv. to current this galvanometer into an ammeter of range 25 a, the required shunt is

Answer 1

Shunt is joined parallel with galvanometer .
Let S is the resistance of shunt , G is the resistance of Galvanometer , Ig is the current r range of Galvanometer, I is the current range of Ammeter
Then formula is given by Ig = IS/(G + S)
Here given, Ig = 15 × 10⁻³ A , I = 25A
G = Vg/Ig = 750mV/15mA = 50Ω

Now, 15 × 10⁻³ = 25S/(50 + S)
⇒0.015 × 50 + 0.015S = 25S
⇒0.75 = 24.985S
S = 0.03Ω

Answer 2

Given :

• v = 750 x 10^-3 v
• Ig = 15 x 10^-3 A
• I = 25 A

To find :

• required shunt

solution :

a = v/I_g...........(formula)

= 750 x 10^-3 v / 15 x 10^-3 A

= 50 ohm

I_g = (S / S + a) x 1 .....(formula)

I_g = 5 / 5 + a x I

15 x 10^-3 = (5 / 5 + 50) / 25

S = 0.03 ohm

=> required shunt is 0.03 ohm