Question

A galvanometer has an internal resistance of 100 Ω and deflects full-scale at 2.00 mA. What size resistor should be added to the galvanometer to convert it to a milliammeter capable of reading up to 4.00 mA, and how should this resistor be connected to the galvanometer?

Answer 1

Answer

The resistor has to be 100

Explanation:

We will have to use the Current Divider Rule, that rule states:

$Ig=\frac{Req}{Rg}*(It)$

where:

Ig= Galvanometer current

It= Total current

Rg= Galvanometer Resistor

Req= Equivalent circuit resistor

For the case of two resistor in parallel:

$Req=\frac{R1*Rg}{R1+Rg}$

now:

$Req=\frac{2mA}{4mA}*100$

Req=50Ω

having the Equivalent resistor we can calculate R1 reformulating the Req formula:

$R1=\frac{(Rg*Req)}{Rg-Req}$

R1=100 Ω

So now when a 4mA current flows into the new circuit, 2mA  will go through the Galvanometer deflecting the full scale.