Question

A 12 ohm galvanometer can show full scale deflection for current of 2.5 ma . How can this galvanometer be used as an ammeter to measure current up to 7.5a?

Answer 1

Explanation:

Given that,

Total current I = 7.5 A

Resistance of a galvanometer G = 12 ohm

Current through galvanometer $I_{g} = 2.5\times10^{-3} A$

We know that,

When the low resistance is connected in parallel with this galvanometer in the circuit then galvanometer works as an ammeter. its low resistance called shunt resistance.  

So, the shunt resistance is

$R_{s} = (\dfrac{I_{g}}{I-I_{g}})\times G$

$R_{s} = (\dfrac{2.5\times10^{-3}\ A}{7.5\ A-2.5\times10^{-3}\ A})\times 12\ ohm$

$R_{s} = 4\times10^{-3}\ ohm$

Hence, the shunt resistance is $R_{s} = 4\times10^{-3}\ ohm$.