Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMOVING CHARGES AND MAGNETISM2 Marks
Question
Explain current sensitivity and voltage sensitivity of galvanometer.
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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.
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