$A$ galvanometer has resistance $G \ \Omega$ and $I_g$ is the current flowing through it which produces full-scale deflection. $S_1$ is the value of the shunt which converts it into an ammeter of range $0$ to $3I$,and $S_2$ is the shunt value which converts it into an ammeter of range $0$ to $4I$. The ratio $S_2:S_1$ is:

  • A
    $\frac{4}{3}$
  • B
    $\frac{3I-I_g}{4I-I_g}$
  • C
    $\frac{3}{4}$
  • D
    $\frac{4I-I_g}{3I-I_g}$

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Similar Questions

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

$A$ galvanometer can be converted into an ammeter by connecting:

Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them,deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is

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To determine the resistance $G$ of a galvanometer by the half-deflection method,a battery of $emf$ $V$ and a series resistance $R$ are used to produce a deflection $\theta$ in the galvanometer. If a shunt resistance $S$ is connected in parallel to the galvanometer to reduce the deflection to $\theta/2$,then $G, R$,and $S$ are related by the equation:

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