Agalvanometer 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

The resistor has to be 100

Explanation:

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

where:

Ig= Galvanometer current

It= Total current

Rg= Galvanometer Resistor

Req= Equivalent circuit resistor

For the case of two resistor in parallel:

now:

Req=50Ω

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

R1=100 Ω

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

the answer is b, have negligible mass

force tends to be low and long diustance masses are high