A galvanometer with a coil resistance of 11.0 and a full-scale current of 0.160 mA?

Question

A galvanometer with a coil resistance of 11.0 and a full-scale current of 0.160 mA is used with a shunt resistor to make an ammeter. The ammeter registers a maximum current of 4.00 mA. Find the equivalent resistance of the ammeter. ____ Ohms

Answer 1

Current through galvanometer is still 0.160A so difference of potenials V = .160*11 = 1.76 mV. But now the full current is I = 4 ma, so effective resistance R = V/I = 0.44 Ohm.

Answer 2

Vmax is the max voltage drop across the galvanometer required to reach a full scale current of 0.160 mA
Vmax=RgImax

Let Im = 4ma (max required current)
then
Vmax=Re Im

so Re=Vmax/Im= 11.0 x 0.160E-3/ 4.00 E-3= 0.440 ohms

Answer 3

The voltage around the galvanometer at finished scale analyzing is 0.17 * 14.0 millivolts (V = IR). The equivalent resistance of the ammeter is hence this voltage divided via 4 mA, which comes out to .595 ohms.