Question
Explain current sensitivity and voltage sensitivity of galvanometer.
✓
Answer
Current Sensitivity :
→For moving coil Galvanometer,
$\phi=\frac{ NBA }{k} \cdot I$
Where, $N =$ number of turns in coil
$A =$ area of coil
$k=$ the torsional constant of spring
$B =$ magnetic field
$I=$ current passing through the coil.
→The deflection per unit current is called the current sensitivity of the galvanometer.
$\therefore \frac{\phi}{ I }=\frac{ BAN }{k}$
→From, equation (2) it is clear that to increase the current sensitivity of galvanometer of coil, number of turns $( N )$ are increased.
Voltage Sensitivity :
→Multiplying $\frac{1}{R}$ on the both sides of equation (2), we get.
$\begin{aligned}
\frac{\phi}{ IR } & =\frac{ BAN }{k R } \\
\therefore \quad \frac{\phi}{ V } & =\frac{ BAN }{k R }
\end{aligned}$
→The deflection per unit voltage is known as the voltage sensitivity of galvanometer.
→Equation (3) shows that if the number of turns in the coil is increased, the resistance of coil also increases, So that the voltage sensitivity remains constant.
→For moving coil Galvanometer,
$\phi=\frac{ NBA }{k} \cdot I$
Where, $N =$ number of turns in coil
$A =$ area of coil
$k=$ the torsional constant of spring
$B =$ magnetic field
$I=$ current passing through the coil.
→The deflection per unit current is called the current sensitivity of the galvanometer.
$\therefore \frac{\phi}{ I }=\frac{ BAN }{k}$
→From, equation (2) it is clear that to increase the current sensitivity of galvanometer of coil, number of turns $( N )$ are increased.
Voltage Sensitivity :
→Multiplying $\frac{1}{R}$ on the both sides of equation (2), we get.
$\begin{aligned}
\frac{\phi}{ IR } & =\frac{ BAN }{k R } \\
\therefore \quad \frac{\phi}{ V } & =\frac{ BAN }{k R }
\end{aligned}$
→The deflection per unit voltage is known as the voltage sensitivity of galvanometer.
→Equation (3) shows that if the number of turns in the coil is increased, the resistance of coil also increases, So that the voltage sensitivity remains constant.
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