Question

A galvanometer has a resistance $G$ and a current $i_{g}$ flowing in it produces full scale deflection. $S_{1}$ is the value of shunt, which converts it into an ammeter of range $0-I$ and $S_{2}$ is the value of shunt for the range $0-2I$ the ratio of $\dfrac{S_{1}}{S_{2}}$ is
\[\begin{align}
  & A.\dfrac{{{S}_{1}}}{{{S}_{2}}}=\left[ \dfrac{2I-{{i}_{g}}}{I-{{i}_{g}}} \right] \\
 & B.\dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{1}{2}\left[ \dfrac{I-{{i}_{g}}}{2I-{{i}_{g}}} \right] \\
 & C.2 \\
 & D.1 \\
\end{align}\]

Answer 1

Hint: We know that a galvanometer is an instrument which can detect the electric current in a circuit. We can convert a moving coil galvanometer into a voltmeter or an ammeter by adding a resistance to the circuit.

Formula: $S=\dfrac{G(i_{g})}{(i-i_{g})}$

Complete answer:
A galvanometer works on the principle of electromagnetic induction, which is when a current carrying conductor is placed in a magnetic field, it experiences a torque.
A galvanometer when connected in parallel to a circuit with low resistance, it can act as an ammeter and measure the current in that circuit. This low resistance is also called the shunt resistance.
Let us assume that a shunt $S$ is connected in parallel with a galvanometer with resistance$G$, then we know that the current in the circuit gets divided. If $i$ is the total current in the circuit and $i_{g}$ is the current in the galvanometer, then the current flowing through the shunt resistance is given as $i_{s}=i-i_{g}$
Since the voltage in both the circuits are equal, we also know from ohms law that $V=IR$.
Applying the same, we get $S(i-i_{g})=G(i_{g})$
Or, $S=\dfrac{G(i_{g})}{(i-i_{g})}$

Here it is given that the maximum current in the ammeter with shunt $S_{1}$ as $0-I$ i.e. $i=I$
Then substituting the values we get, $S_{1}=\dfrac{G(i_{g})}{(I-i_{g})}$
Also given that, the maximum current in the ammeter with shunt $S_{2}$ is$0-2I$i.e. $i=2I$
Then substituting the values we get, $S_{2}=\dfrac{G(i_{g})}{(2I-i_{g})}$
But we need the ratio$\dfrac{S_{1}}{S_{2}}$
Then taking the ratio between $\dfrac{S_{1}}{S_{2}}$ we get, $\dfrac{S_{1}}{S_{2}}=\dfrac{\dfrac{G(i_{g})}{(I-i_{g})}} {\dfrac{G(i_{g})}{(2I-i_{g})}}=\dfrac{(2I-i_{g})}{(I-i_{g})}$

Hence the answer is \[A.\dfrac{{{S}_{1}}}{{{S}_{2}}}=\left[ \dfrac{2I-{{i}_{g}}}{I-{{i}_{g}}} \right]\]

Note:
We know that the sensitivity of a galvanometer is its ability to give large deflections even for a small value of the current. And it is given as the ratio between the change deflections in the galvanometer to the change in current in the coil.