Answer:
Each ammeter in the loops would read 4A and if other loop is added the loop ammeter would read 3A
Explanation:
The formula for current in parallel is;
[tex]I_{total} = I_{1} + I_{2} + I_{3} ...[/tex] [tex]+ I_{n}[/tex]
The total current value from the source battery is 12A, so the three loops with no load would read;
= 12 / 3 = 4A
When another loop is added to the circuit, the total current would be divided by the four loop which will give a value of 3A for each loop.