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Question: Answered & Verified by Expert
A 50 Ω resistance is connected to a battery of 5 V. A galvanometer of resistance 100 Ω is to be used as an ammeter to measure current through the resistance, for this a resistance rS is connected to the galvanometer. Which of the following connections should be employed if the measured current is with in 1% of the current without the ammeter in the circuit?
PhysicsCurrent ElectricityJEE MainJEE Main 2016 (09 Apr Online)
Options:
  • A rS=0.5Ω in series with galvanometer
  • B rS=1 Ω in series with galvanometer
  • C rS=1 Ω in parallel with galvanometer
  • D rS=0.5 Ω in parallel with the galvanometer
Solution:
1514 Upvotes Verified Answer
The correct answer is: rS=0.5 Ω in parallel with the galvanometer

Initially current in the circuit is 

I=550=0.1

and the current in circuit when ammeter is connected,is 1 % of initial current , i.e.,  I=1100× 0·1 = 0.099 A

To increase the range of galvanometer, let resistance rs
 is connected parallel to it and along with 50 Ω is connected in series to ammeter. So, equivalent resistance  of circuit is:

Req=50+100 rS100+rS

Now, From equation V= I' Req

5 = 0.099×50+100 rs100+rs50+100 rs100+rs = 50.09950+100 rs100+rs =50.5050100 rs100+rs = 50.5050rs = 0.5 Ω

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