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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?
- Looking at small germs.
- Looking at circular spots.
- Looking at a vertical tube containing some water.
A $15.0 \mu F$ capacitor is connected to a $220 V , 50 Hz$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?
→The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point?
Figure shows an electrolyte of AgCl through which a current is passed. It is observed that 2.68g of silver is deposited in 10 minutes on the cathode. Find the heat developed in the $20\Omega$ resistor during this period. Atomic weight of silver is 107.9g/mole.
→The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32cm below the top surface of the slab. Calculate the dispersive power of the material.
Write the definition of electric current and ampere $(A)$ and different units of electric current.
→Explain nuclear energy.
Draw the graph of variation of Coulomb force (F) with $(r)$. Where the distance between the two charges of each charge pair is ( $1 \mu C , 2 \mu C$ ) and ( $2 \mu C$, $-3 \mu C$ ). Analyse the graph obtained.
→The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be $2\ cm$ below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?
→The energy gaps in the energy band diagrams of a conductor, semiconductor and insulator are $E_1, E_2$ and $E_3.$ Arrange them in increasing order.
→