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

The resistor has to be 100

Step-by-step explanation:

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

`Ig=(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=(R1*Rg)/(R1+Rg)`

now:

`Req=(2mA)/(4mA)*100\\`

Req=50Ω

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

`R1=((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.