Q157 / 180 · COMEDK 2017 · Physics

In an ammeter $10 \%$ of main current is passing through the galvanometer. If the resistance of the galvanometer is $G$, then the shunt resistance (in $\Omega$ ) is

  • A
    $9 G$
  • Correct answer
    B
    $\frac{G}{9}$
  • C
    $90 G$
  • D
    $\frac{G}{90}$

Solution

If $I$ be the main current, then current passing through galvanometer, $I_{g}=10 \% \text { of } I=\frac{I}{10}$ Resistance of galvanometer, $R_{g}=G$ Shunt resistance, $R_{S}=\frac{I_{g} R_{g}}{I-I_{g}}=\frac{\frac{I}{10} \times G}{I-\frac{I}{10}}=\frac{G}{9}$

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

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