In the given diagram,the cell and the ammeter, both have negligible resistance. the resisitors are identical. With switch K open , the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed?
When the switch is open, the effective resistance is R/2, since the two resistances R each are in parallel.
Therefore, the pd, V = I x R/2 = 0.6 x R/2 = 0.3 R
When the key is closed, the effective resistance becomes R/3
Now the reading of the ammeter, I’ = V/(R/3) = 0.3R/(R/3)=0.9A