Q. 35 B4.2( 26 Votes )

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?

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


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
What is Electric Current?56 mins
Understanding Short Circuit38 mins
Ohm's Law43 mins
Factors Affecting Resistance45 mins
Quiz | Most Challenging Questions on Electricity40 mins
What is Power Rating?38 mins
Genius Quiz | Electricity - Test your Speed and Accuracy38 mins
Parallel combination of resistors36 mins
Lets Understand Fleming's Left Hand Rule39 mins
Quiz | Rapid Fire Round (Electricity)44 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses