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AP EAMCET202123 Aug 2021Morning ShiftPhysicsCurrent ElectricityActual

A galvanometer of resistance 40 gives a deflection of 10 divisions per mA . There are 50 divisions on the scale. Maximum current that can pass through it when a shunt resistance of 2 is connected is

Options

  1. A105 ~mA
  2. B155 ~mA
  3. C210 ~mA
  4. D75 ~mA

Correct answer

A. 105 ~mA

Step-by-step solution

Given, galvanometer resistance, R_G=40 Shunt resistance, R_ =2 Reading =10 div / mA and number of divisions, n=50 Galvanometer current, I_G= 50 10 =5 ~mA Let shunt current be I . array ll Since, I R_G+R_ & = I_G R_ I & = 5 (40+2) 2 & =105 ~mA array

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