Question

You are supplied with a galvanometer, resistor, and some connecting wires.

1. Using a circuit diagram, show how will you convert the given galvanometer into an ammeter. 

2. Find the expression for the shunt resistance in the circuit.

3. A galvanometer is to be converted into an ammeter of range 0 – 1 A. Galvanometer has resistance 100Ω and the current for full scale deflection is 10 mA. Find the length of the nichrome wire to be used as shunt.

Given, Resistivity ρ = 1.1 x 10^-6 Ωm

Diameter of the wire = 1 mm

Answer 1

2. Let I_g be the current through the galvanometer of resistance R_G and the shunt resistance be r_s .

Let I be the current to be measured by the converted ammeter.

We can write,

I_gR_G = \((I-I_g)r_s\)

∴ r_s = \(\frac{I_gR_G}{(I-I_g)}\)

3. Given I = 1A

I_g = 10 mA

R_G = 100Ω

Diameter = 1 mm

∴ r_s = \(\frac{I_gR_G}{(I-I_g)}\) = \(\frac{10\times10^{-3}\times100}{(1-0.01)}\) = 1Ω

r_s = \(\frac{ρl}{\pi r^2}\)

I = \(\frac{r_s\pi r^2}{ρ}\) = \(\frac{1\times3.14\times(0.50\times10^{-3})^2}{1.1\times10^{-6}}\) = 0.714 m