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A 20 ohm shunt is joined with an ammeter with 200 ohm resistance. It is joined to a 10v battery having a resistance of 4 ohm. What will the observation of the ammeter be?

The circuit diagram of the problem is as given below. Ammeter shunt value is as shown in the circuit. The ammeter shunt is used to divert most of the circuit current towards shunt resistance and fraction of current towards the ammeter.

Total resistance of the circuit(R)

= 4+(200 x 20/220)

= 4+ 18.18

R=22.18 Ohms

Total current of the circuit

I = V/R= 10/22.18=0.45 Amps

The current through ampere meter

= 0.45 x 20/220b =0.0409 Amps.

= 40.9 mA