Define the current sensitivity of a galvanometer. Write its SI un… - Vidyadip

Question

Define the current sensitivity of a galvanometer. Write its $SI$ unit.
Figure shows two circuits each having a galvanometer and a battery of $3 V$ .
When the galvanometer in each arrangement do not show any deflection, obtain the ratio $R_1 / R_2$.

✓

Answer

Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.
The $SI$ unit of current sensitivity is rad. $A ^{-1}$. Current sensitivity is expressed as $\frac{\theta}{I}=\frac{N A B}{K}$ where $N , A , B$ and $K$ are number of turns, cross $-$ sectional area, magnetic field intensity and galvanometer's constant respectively.

For balanced Wheatstone bridge, there will be no deflection in the galvanometer.
$\therefore \frac{4}{R_1}=\frac{6}{9}$
$\Rightarrow R_1=\frac{4 \times 9}{6}=6 \Omega$

For the equivalent circuit. when the Wheatstone bridge is balanced, there will be no deflection in the galvanometer.
$\therefore \frac{12}{8}=\frac{6}{R_2}$
$\Rightarrow R_2=\frac{6 \times 8}{12}=4 \Omega$
$\therefore \frac{R_1}{R_2}=\frac{6}{4}=\frac{3}{2}$

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