Question

Define the current sensitivity of a galvanometer. Write S.I unit
Figure shows two circuits each having a galvanometer and a battery of 3V. When the galvanometer in each arrangement do not show any deflection obtain the ratio ${\displaystyle \frac{R_1}{R_2}}$

Answer 1

Hint- In order to solve this question, we will use the concept of wheatstone bridge. In a wheatstone bridge, the bridge is said to be balanced when the galvanometer shows zero deflection. We will use the condition of zero deflection of the galvanometer to find the unknown value of the resistance and then proceed accordingly.

Complete step-by-step answer:
Condition for balanced bridge-

${\displaystyle \frac{R_1}{R_3}}={\displaystyle \frac{R_2}{R_4}}$

Current sensitivity- The term current sensitivity is defined as when current flow through the galvanometer coil, the coil shows deflection.

Mathematically it can be represented as

$I_s={\displaystyle \frac{\theta }{T}}={\displaystyle \frac{nBA}{c}}$

Where n is the number of turns in the coil, B is the magnetic field around the coil, A is the cross sectional area of the coil and c is the restoring torque.

The SI unit of the current sensitivity is radian per ampere.

Now for calculating the value of unknown resistance we will use the condition of balanced bridge.

First for the value of resistance $R_1$

${\displaystyle \frac{R_x}{R_3}}={\displaystyle \frac{R_2}{R_4}}$
Where $R_x=4,R_2=R_{1,}{R}_3=9,R_4=6$

Substituting these values in the above equation

$$\begin{gathered} {\displaystyle \frac{R_1}{9}}={\displaystyle \frac{4}{6}} \\ {R}_1=6\mathrm{\Omega } \end{gathered}$$

Now for the value of resistance $R_2$ from figure second

${\displaystyle \frac{R_1}{R_3}}={\displaystyle \frac{R_2}{R_x}}$
Where $R_1=6,R_2=12,R_3=8,R_x=R_2$

Substituting these values in the above equation

$$\begin{gathered} {\displaystyle \frac{6}{R_2}}={\displaystyle \frac{12}{8}} \\ {R}_2={\displaystyle \frac{6\times 8}{12}}=4\mathrm{\Omega } \end{gathered}$$

Therefore the ratio of ${\displaystyle \frac{R_1}{R_2}}$ is
${\displaystyle \frac{R_1}{R_2}}={\displaystyle \frac{6}{4}}={\displaystyle \frac{3}{2}}$

Note- In order to solve questions related to the wheatstone bridge, learn the basic concept behind the wheatstone bridge and also remember the wheatstone bridge is used to find the value of the unknown resistance. It is used to measure the value of the resistance very precisely. Other applications of the wheatstone bridge is ; it is used with operational amplifiers to measure temperature, light, strain etc.