Question

A galvanometer has a coil of resistance 41 and gives f.s.d with a current of 10mA. How may it be made into an ammeter reading up to 10A?​

Answer 1

Answer:

Essentially you need to add a shunt resistance in parallel with the galvanometer movement, which is sized to cause 10mA to flow through the galvanometer movement, when 10A is flowing through the parallel combination of the galvanometer movement and the shunt resistor, with the difference in current flowing through the shunt. The problem is calculating the value of that shunt resistance.

The solution involves applying Kirchoff’s Current Law (KCL), and Ohm’s Law for a current divider network, where Rm is the resistance of the galvanometer movement, and Rs is the value of the needed shunt resistance.

Substituting your specified conditions (It = 10A, I'm = 10mA, and Rm = 50Ω) into the relationships above, we get the following results:

The problem, of course, is coming up with that value of resistance (which is not only extremely small, but extremely precise), and having a resistance-measuring instrument of sufficient precision to measure that resistance, to confirm it’s the required value. As you can see, the smaller the current necessary to cause full-scale deflection of the meter (I'm), and the larger the total current (It) you’re trying to measure, the smaller the shunt resistor (Rs) is going to be, and the harder it’s going to be to construct it and verify that it is the required value. In reality, Rs is probably going to be a short piece of wire connected across the galvanometer terminals, but given the value required, the accuracy of the finished meter + shunt is going to be dependent on things like how tightly you secure the hardware on the meter lugs, oxidation on the wire surfaces, etc.

Fortunately, you can buy ammeter shunts. If you come up with one that’s close in value, you can file it to alter (increase) its resistance.