Maximum current that an ammeter can read?

Question

Here is the question:

The coil of a galvanometer has a resistance of 20.0 ohms, and its meter deflects full scale when a current of 6.20 mA (mili-amperes) passes through it. To make the galvanometer into an ammeter, a 24.8 m-omhs (mili-ohms) shunt resistor is added to it. What is the maximum current that this ammeter can read?

The back of the book gives the answer as 5.01 amperes, but I am not sure how they obtained this result. It would be very helpful if someone could work this problem out step-by-step and show me how to do this type of problem.

Answer 1

At a full scale reading, the coil is carrying 6.20mA. Its resistance is 20ohms, so the voltage across the coil is 6.2*20 millivolts. The same voltage is across the shunt resistor, which is 24.8 m-ohms. The current through the shunt is then 6.2*20/24.8 amps = 5 amps. Add this to the 0.0062A in the coil and you get 5.0062 amps, or round to 5.01 amps.

Answer 2

The shunt resistor is placed in parallel to the galvonometer. Therefore, the voltage drop across both is the same. So, using V = IR on both galvonometer and shunt and setting them equal:

(20.0 ohms)(galv current) = (24.8 m-ohms)(shunt current)

shunt current = (galv current) (20 / 0.0248)

Total current = galv current (1 + 20 / 0.0248)

Plug in the max current of 6.20 milli amps into that equation to get the maximum total current.

Answer 3

Galvanometer Into Ammeter