An analog ammeter is made by a galvanometer (with certain \$i_{max}\$ and resistance \$r\$), in parallel with a shunt resistor \$R_s\$ that depends upon the max scale chosen.

Nevertheless, when an ammeter is concerned, we talk about a series resistance of the ammeter. Is this series resistance simply the parallel equivalent of \$R_{s}\$ and \$r\$? that is

$$R_{series}=\frac{R_{s} r}{r+R_{s}}$$

  • \$\begingroup\$ yes but the parallel r across the voltage meter is far smaller by orders of magnitude \$\endgroup\$ Mar 24, 2017 at 3:04

1 Answer 1


Full answer: Yes :-).

In some cases a series resistor may be added in series with the meter to help adjust values to standard ones or to calibrate, but this degrades the overall performance and is non ideal.


Meter current full scale = Im consider as imax or imeter )
Full scale of overall meter system = Ir = irange
Rmeter = Rm
Rshunt = Rparallel = Rp (avoiding Rs with "series" connotation)

The current through the meter = Im.
The current through the shunt = (Ir-Im)
As they are in parallel the resistances are inversely proportional to their currents.
So Rp = Rm x Im / (Ir - Im)


Rm = 1000 Ohms
Im = 50 uA
Ir = 200 mA
Rshunt = Rp = 1000 Ohms x 50 uA/ (200 mA - 50 uA)
Rp = 1000 x 50 / 199,950 = 0.2500625 Ohms
I've calculated Rp to so many significant figures to demonstrate how close to 0.25 Ohms it is.

ie if the resistors are simply scaled in the ratios of the load current and meter current we get
Rp = Rm x 50/200000 = 0.25 Ohms.
In this case Ir/Im = 200000 uA/50 uA = 4000:1 and the fact that the meter current is subtracted from the load current is hardly noticeable in calculating values. For Im closer to Ir the difference needs to be accounted for.


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