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Question

A galvanometer with a full-scale sensitivity of 2x10^-3A has a coil resistance of 100ohms. If it is to be used in an anmeter with a full-scale reading of 3.0A, what is the necessary shunt resistance?

Answer 1

Voltage across galvanometer coil is E = IR = 0.002 x 100 = 0.2 volts.
Voltage across the parallel shunt is the same.
Current in the shunt is 3.000 – 0.002 = 2.998 amperes.
Resistance of shunt is E/I = 0.2 / 2.998 = 0.06671 ohms, to four significant figures.

This is virtually the same value you would get if you assumed ALL the current passed through the shunt, i.e. 0.2 / 3 = 0.6667, a difference of about 44.47 micro ohms.

Since the current through the shunt is three orders of magnitude greater than the galvanometer current., and it is difficult to read a galvanometer to better than 1%, in practice you would use a shunt resistance of 0.067 ohms. Practically, it is very difficult to measure such a small resistance accurately, much less attach it to the terminal posts of the galvanometer without the connection affecting the shunt resistance. Also, the coil resistance of the galvanometer changes slightly with temperature so even a theoretically correct shunt resistor will only be correct at one temperature. Worse, unless you use a shunt resistor with a low temperature coefficient of resistance, the shunt will change temperature due to self-heating.

The easiest way to install a shunt is to wrap a short length of wire with a low temperature coefficient of resistance around both terminals of the galvanometer, leaving enough slack to adjust the length of the wire between the terminals. Connect the galvanometer and its shunt in series with a ammeter of known accuracy, a variable dc power supply, and a current limiting resistor of about one ohm or so. Adjust the power supply to produce 3A current on the meter of known accuracy. Adjust the length of the shunt between the galvanometer terminals to read full scale. Turn off the power supply (or open the circuit) when making adjustments to avoid toasting the galvanometer coil.

Answer 2

Do you understand what is being asked?
The shunt resistor carries the bulk of the current so that a small, sensitive meter can be used.

At full scale, the voltage across the galvanometer must be Vg. (How would you figure that out?)

Full-scale on the galvanometer corresponds to 3 A current in the main circuit. Therefore Vg is the voltage drop across the parallel resistances Rs and Rg (Rg = galvanometer, Rs = shunt) when the current is 3 A.

So you can find the parallel resistance Rp = (1/Rg + 1/Rs)^-1 and Rp = Vg / 3A.

Now since you are given Rg, you can figure out Rs.

Here's a question back at you: how much error would be incurred if you ignored Rg when calculating the shunt resistance? (i.e. treat Rg as infinite; then Rp = Rs.)

A more direct way to approach the problem, which might be a little difficult to understand, is to note that the main current, 3 A = It, is the sum of Ig and Is (galvanometer and shunt currents) and that, since they have the same voltage drop, the ratio Ig / Is is the same as the ratio Rs / Rg.
Now if It = Is + Ig
Is = Ig *Rg/Rs
It = 3
Ig = 0.002
Rg = 100
you can solve that for Rs.

Answer 3

The current in the shunt will be 3/0.002 times the current through the coil, or 1500 times. Thus the shunt resistance must be 100/1500 ohms.