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

Question

Philoid · Accepted Answer

When switch is open, the upper two resistances are connected in parallel in the circuit.

Effective resistance is 1/Req=1/R+1/R=2/R

Req = R/2

So the current=I=V/(R/2)=0.6A (given)

V/R = 0.3 A

When the switch closes, the third resistance also comes in the circuit.

The effective resistance of the circuit becomes R/3

Hence, Current I = V/(R/3) = 3 (V/R) = 3 x 0.3 = 0.9 A

In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identicalWith the switch K open, the ammeter reads 0.6 A. What will the ammeter reading when the switch is closed?

In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical

With the switch K open, the ammeter reads 0.6 A. What will the ammeter reading when the switch is closed?