Question

A ammeter is to be constructed using a galvanometer that has a 2.00 mA full-scale deflection and internal resistance of 50.0 ohms. It must be good enough to detect 40 mA, which is the max current. How would one do this? How would you make a voltmeter able to read the largest voltage drop(which is 14.0 v) in the circuit?

Please show work. Thank You

Answer 1

Va = Ia*Ra = 2*10^-3 * 50 = 0.10 Volts =
Voltage across ammeter @ full-scale.

Rp=Va/I = 0.1/(40-2)*10^-3=0.1/38*10^-3=
2.632 Ohms. = Resistor connected in
parallel with the ammeter to extend
current range to 40 mA.

Voltmeter:

Rs = (V-Va)/If = (14-0.10)/2*10^-3=6,950
Ohms = The resistor connected in series
with the ammeter.

Note: The 2.632-Ohm resistor is not used
with the voltmeter arrangement.