(A) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.Given: $G = 25\,\Omega$ and $S = 2.5\,\Omega$.Let $I$ be the total c

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(A) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.Given: $G = 25\,\Omega$ and $S = 2.5\,\Omega$.Let $I$ be the total current and $I_g$ be the current flowing through the galvanometer.According to the current divider rule,the current through the galvanometer is given by:$I_g = I \times \frac{S}{G + S}$Therefore,the ratio of the current through the galvanometer to the total current is:$\frac{I_g}{I} = \frac{S}{G + S}$Substituting the given values:$\frac{I_g}{I} = \frac{2.5}{25 + 2.5} = \frac{2.5}{27.5}$$\frac{I_g}{I} = \frac{25}{275} = \frac{1}{11}$

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

Medium

We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

A
$\frac{I_g}{I} = \frac{1}{11}$
B
$\frac{I_g}{I} = \frac{1}{10}$
C
$\frac{I_g}{I} = \frac{3}{11}$
D
$\frac{I_g}{I} = \frac{4}{11}$

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Moving Charges and Magnetism

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

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

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Medium

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MediumMHT CET 2008

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When an additional resistance $1980 \ \Omega$ is connected in series with a voltmeter,each scale division has $100$ times larger value. The resistance of the voltmeter is (in $\Omega$)

EasyMHT CET 2022

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$A$ galvanometer has a resistance of $25 \, \Omega$ and a maximum of $0.01 \, A$ current can be passed through it. In order to change it into an ammeter of range $10 \, A$, the shunt resistance required is

Medium

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We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

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