(D) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 25 \times 4 \time

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(D) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 25 \times 4 \times 10^{-4} \, A = 100 \times 10^{-4} \, A = 0.01 \, A$. To convert a galvanometer into a voltmeter, a high resistance $R$ must be connected in series with it. The formula for the series resistance is $R = \frac{V}{I_g} - G$, where $V = 25 \, V$, $I_g = 0.01 \, A$, and $G = 50 \, \Omega$. Substituting the values: $R = \frac{25}{0.01} - 50 = 2500 - 50 = 2450 \, \Omega$. Therefore, a resistance of $2450 \, \Omega$ should be connected in series.

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

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$A$ galvanometer of $50 \, \Omega$ resistance has $25$ divisions. $A$ current of $4 \times 10^{-4} \, A$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25 \, V$, it should be connected with a resistance of

AIPMT 2004 All AIPMT

A
$2500 \, \Omega$ as a shunt
B
$2450 \, \Omega$ as a shunt
C
$2550 \, \Omega$ in series
D
$2450 \, \Omega$ in series

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Moving Charges and Magnetism

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

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$A$ galvanometer of $50 \, \Omega$ resistance has $25$ divisions. $A$ current of $4 \times 10^{-4} \, A$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25 \, V$, it should be connected with a resistance of

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In the given figure,an ammeter $A$ consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is . . . . . . $mA$.

In an ammeter,$4 \%$ of the main current is passing through the galvanometer. If the shunt resistance is $5 \Omega$,then the resistance of the galvanometer will be: (in $Omega$)

$A$ $36 \,\Omega$ galvanometer is shunted by a resistance of $4 \,\Omega$. The percentage of the total current that passes through the galvanometer is (in $\%$)

To verify Ohm's law,a student is provided with a test resistor $R_T$,a high resistance $R_1$,a small resistance $R_2$,two identical galvanometers $G_1$ and $G_2$,and a variable voltage source $V$. The correct circuit to carry out the experiment is

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