(A) Current sensitivity $({\sigma _i})$ is defined as the deflection per unit current: ${\sigma _i} = \frac{\theta}{I}$.Voltage sensitivity $({\sigma

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$\frac{{\sigma _i}}{G} = {\sigma _V}$

(A) Current sensitivity $({\sigma _i})$ is defined as the deflection per unit current: ${\sigma _i} = \frac{\theta}{I}$.Voltage sensitivity $({\sigma _V})$ is defined as the deflection per unit voltage: ${\sigma _V} = \frac{\theta}{V}$.Since $V = I \cdot G$,where $G$ is the resistance of the galvanometer,we can substitute $V$ in the voltage sensitivity equation:${\sigma _V} = \frac{\theta}{I \cdot G}$.Substituting ${\sigma _i} = \frac{\theta}{I}$ into the expression for ${\sigma _V}$,we get:${\sigma _V} = \frac{{\sigma _i}}{G}$.Therefore,the relation is $\frac{{\sigma _i}}{G} = {\sigma _V}$.

Class 12 Physics Moving Charges and Magnetism The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion

Easy

The relation between voltage sensitivity $({\sigma _V})$ and current sensitivity $({\sigma _i})$ of a moving coil galvanometer is (Resistance of Galvanometer = $G$)

A
$\frac{{\sigma _i}}{G} = {\sigma _V}$
B
$\frac{{\sigma _V}}{G} = {\sigma _i}$
C
$\frac{G}{{\sigma _V}} = {\sigma _i}$
D
$\frac{G}{{\sigma _i}} = {\sigma _V}$

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Moving Charges and Magnetism

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The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion

$A$ galvanometer having $30$ divisions has a current sensitivity of $0.0625 \frac{\text{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of $6 \text{ V}$,then the resistance of that voltmeter is:

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$A$ galvanometer having a coil resistance of $30\,\Omega$ shows full-scale deflection when a current of $2\,A$ passes through it. It can be converted into an ammeter to read currents up to $10\,A$ by:

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The scale of a galvanometer is divided into $25$ equal divisions. The resistance of the galvanometer is $100 \,\Omega$. The current sensitivity of the galvanometer is $4 \times 10^{-4} \,A/div$. What resistance in $ohm$ must be connected in series to measure $2.5 \,V$?

Medium

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Which one of the following modifications may increase the sensitivity of a moving coil galvanometer?
$1^{st}$ Way: By using a spring of smaller torsion constant.
$2^{nd}$ Way: By using a smaller coil.
$3^{rd}$ Way: By using a stronger magnet.
$4^{th}$ Way: By using a coil having fewer number of turns.

Medium

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$A$ galvanometer coil has a resistance of $10 \ \Omega$ and the meter shows full-scale deflection for $3 \ \text{mA}$. The value of the shunt required to convert this meter into an ammeter of range $0$ to $10 \ \text{A}$ is . . . . . . $\Omega$.

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The relation between voltage sensitivity $({\sigma _V})$ and current sensitivity $({\sigma _i})$ of a moving coil galvanometer is (Resistance of Galvanometer = $G$)

Similar Questions

$A$ galvanometer having $30$ divisions has a current sensitivity of $0.0625 \frac{\text{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of $6 \text{ V}$,then the resistance of that voltmeter is:

$A$ galvanometer having a coil resistance of $30\,\Omega$ shows full-scale deflection when a current of $2\,A$ passes through it. It can be converted into an ammeter to read currents up to $10\,A$ by:

The scale of a galvanometer is divided into $25$ equal divisions. The resistance of the galvanometer is $100 \,\Omega$. The current sensitivity of the galvanometer is $4 \times 10^{-4} \,A/div$. What resistance in $ohm$ must be connected in series to measure $2.5 \,V$?

Which one of the following modifications may increase the sensitivity of a moving coil galvanometer?$1^{st}$ Way: By using a spring of smaller torsion constant.$2^{nd}$ Way: By using a smaller coil.$3^{rd}$ Way: By using a stronger magnet.$4^{th}$ Way: By using a coil having fewer number of turns.

$A$ galvanometer coil has a resistance of $10 \ \Omega$ and the meter shows full-scale deflection for $3 \ \text{mA}$. The value of the shunt required to convert this meter into an ammeter of range $0$ to $10 \ \text{A}$ is . . . . . . $\Omega$.

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