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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 r S 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?

Options

  1. Ar S = 0.5 Ω in series with galvanometer
  2. Br S = 1 Ω in series with galvanometer
  3. Cr S = 1 Ω in parallel with galvanometer
  4. Dr S = 0.5 Ω in parallel with the galvanometer

Correct answer

D. r S = 0.5 Ω in parallel with the galvanometer

Step-by-step solution

Initially current in the circuit is I = 5 50 = 0.1 and the current in circuit when ammeter is connected,is 1   % of initial current , i.e., I ′ = 1 100 ×   0 · 1   =   0.099   A To increase the range of galvanometer, let resistance r s is connected parallel to it and along with 50   Ω is connected in series to ammeter. So, equivalent resistance of circuit is: R e q = 50 + 100   r S 100 + r S Now, From equation V =   I '   R e q 5   =  

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