(C) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 50 \times 2 \time

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(C) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 50 \times 2 \times 10^{-4} \text{ A} = 10^{-2} \text{ A}$.To convert a galvanometer into a voltmeter,a high resistance $R_H$ is connected in series with it.The formula for the series resistance is $R_H = \frac{V}{I_g} - R_g$,where $V = 25 \text{ V}$ and $R_g = 25 \Omega$.Substituting the values: $R_H = \frac{25}{10^{-2}} - 25 = 2500 - 25 = 2475 \Omega$.

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

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$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

A
$1200$
B
$1225$
C
$2475$
D
$2500$

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Class 12

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Moving Charges and Magnetism

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

$A$ galvanometer has a $50$ division scale. The battery has no internal resistance. It is found that there is a deflection of $40$ divisions when $R = 2400\,\Omega$. The deflection becomes $20$ divisions when the resistance taken from the resistance box is $4900\,\Omega$. Then we can conclude:

DifficultJEE MAIN 2016

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$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

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To send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$,the shunt required is . . . . . . . (in $\Omega$)

EasyGUJCET 2011

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The current passing through the galvanometer is $4\%$ of the total current in the circuit. If the resistance of the galvanometer is $G$, the shunt resistance $S$ connected across the galvanometer is:

DifficultMHT CET 2026

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$A$ moving coil galvanometer of resistance $100 \ \Omega$ shows full-scale deflection when a current of $100 \ \mu A$ passes through it. If it is intended to show full-scale deflection when a current of $1 \ mA$ passes through it,the value of shunt resistance in ohms to be connected to the galvanometer is:

DifficultTS EAMCET 2002

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$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

Similar Questions

$A$ galvanometer has a $50$ division scale. The battery has no internal resistance. It is found that there is a deflection of $40$ divisions when $R = 2400\,\Omega$. The deflection becomes $20$ divisions when the resistance taken from the resistance box is $4900\,\Omega$. Then we can conclude:

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

To send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$,the shunt required is . . . . . . . (in $\Omega$)

The current passing through the galvanometer is $4\%$ of the total current in the circuit. If the resistance of the galvanometer is $G$, the shunt resistance $S$ connected across the galvanometer is:

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