im doing a lab experiment on dc meters using galvanometer to convert to ammeter w/diff resistance need theory?

Question

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Answer 1

You need to know the resistance of the galvanometer. If not specified, determine it by measuring the voltage across the galvanometer at a given current reading. Then the galvanometer is converted to an ammeter by using a resistance across the galvanometer terminals. You compute the amp scale by using the fact that the input current will divide into two paths: one through the galvanometer, and one through the parallel resistor. The ratio of the two path's currents is inverse to the resistance values. For example, if the parallel resistor equals the galvanometer resistance, then twice the galvanometer reading represents the input current.

Let Rg = galv resistance, Rp = shunt resistance, Ig = galv current (indicated on the galv), and Ip = current through shunt. Then

Ig*Rg = Ip*Rp (the voltage across both resistances is the same)

Ip = Ig*Rg/Rp

The total current It into the circuit is Ig+Ip, so

It = Ig + Ig*Rg/Rp

It = Ig(1+Rg/Rp)

If Rg = Rp, then It = 2*Ig

One problem with this method is that if Rg is very low, then very very low values of Rp are needed to get high-amp scales. This can be overcome by adding a resistor Rs in series with the galvanometer, so that Rg' becomes Rg + Rs.
However, doing this increases the overall impedance of the meter, increasing its effect on any circuit it is used to measure.