Q124 / 170 · Manipal (MET) 2015 · Physics

We have a galvanometer of resistance $25 \Omega$. It is shunted by $2.5 \Omega$ wire. The part of the total current that flows through the galvanometer is given as

  • A
    $\frac{i}{i_0}=\frac{4}{11}$
  • B
    $\frac{i}{i_0}=\frac{3}{11}$
  • C
    $\frac{i}{i_0}=\frac{2}{10}$
  • Correct answer
    D
    $\frac{i}{i_0}=\frac{1}{11}$

Solution

When a galvanometer is connected to the shunt resistance, then we have potential drop across the galvanometer = potential drop across the shunt i.e. $i G=\left(i-i_0\right) S$ ...(i) Here, $i=$ current through the galvanometer $G=$ resistance of the galvanometer $i_0=$ current through the shunt resistance $S=$ shunt resistance From Eq. (i), we have the part of the total current that flows through the galvanometer is $\frac{i}{i_0}=\frac{S}{G+S}$ Substituting, $S=2.5 \Omega, G=25 \Omega$, we get $\frac{i}{i_0}=\frac{1}{11}$

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

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