In the given circuit diagram, the cell and the ammeter, both have negligible resistance. The resistances are identical. With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed?

Explanation

Let the value of each resistance be .R..
(i) When key .K. is open the only two resistance are in circuit so resultant resistance in this case is parallel combination of two resistances,
`(1)/(R_(1))= (1)/(R)+ (1)/(R)= (2)/(R)`
`implies R_(1)= (R)/(2)`

Let potential difference of cell = V volt
Current I = 0.6 A (given)
`:. V= IR_(1)= 0.6 xx (R)/(2)= 0.3R`

(ii) Now, when key .K. is closed, all three resistances are parallel in the circuit. Therefore,
`(1)/(R_(2))= (1)/(R)+(1)/(R)+(1)/(R)= (3)/(R)`
`implies R_(2)= (R)/(3)`
Using Ohm.s law `V= IR_(2)`
`0.3R= I((R)/(3))`
`implies I= 0.3 R xx (3)/(R)= 0.9A`
Therefore, the ammeter reads 0.9 A.