Q146 / 180 · COMEDK 2018 · Physics

A certain current on passing through a galvanometer produces a deflection of 100 divisions. When a shunt of one ohm is connected, then the deflection reduces to 1 division. The galvanometer resistance is

  • A
    $100 \Omega$
  • Correct answer
    B
    $99 \Omega$
  • C
    $10 \Omega$
  • D
    $9.9 \Omega$

Solution

Given, shunt resistance, $R_{S}=1 \Omega$ Total current, $I=100 \mathrm{~A}$ Galvanometer current, $I_{g}=1 \mathrm{~A}$ We know that, $$ \begin{array}{ll} & I_{g}=\frac{I R_{\mathrm{S}}}{R_{\mathrm{S}}+G} \\ \Rightarrow & 1=\frac{100 \times 1}{1+G} \\ \Rightarrow & 1+G=100 \\ \Rightarrow & G=99 \Omega \end{array} $$

Hence, the correct answer is option: 'B'.

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