Question

In the circuit diagrams ($A,B,C$ and $D$) shown below, $R$ is a high resistance and $S$ is a resistance of the order of galvanometer resistance $G$. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as :

• A. Circuit A with G = $\frac{RS}{\left(R-S\right)}$
• B. Circuit B with G = S
• C. Circuit C with G = S
• D. Circuit D with G = $\frac{RS}{R-S}$

Answer 1

Answer: (D)

Here, in figure $D$, current will flow through the circuit when key $K_1$ is closed and $K_2$ is open. The current flowing through the galvanometer is proportional to the deflection in it.
$I_1=\frac{E}{R+G}=k\theta$
where, $E$ - emf of the cell
$R$ - resistance from the resistance box
G- galvanometer resistance for current I
$\theta$ - galvanometer deflection for current I
$k$ - proportionality constant.
When $K_2$ is closed and by adjusting the shunt resistance $S$, we can make galvanometer deflection as $\theta /2$.
Then the current in the circuit is ;
$I_2=\frac{E}{R+\frac{GS}{G+S}}$
Now, a fraction, $\left(\frac{S}{G+S}\right)$ of the current in the circuit is flows through the galvanometer, which is given by:
$I^{\prime }=\frac{I_{2}S}{G+S}=k\theta /2$
Now, from the above relations, we can get the resistance of the given galvanometer as:
$G=\frac{RS}{R-S}$