(D) First, calculate the resistance of the galvanometer $(G)$: $G = \frac{V_g}{I_g} = \frac{0.75 \text{ V}}{10 \times 10^{-3} \text{ A}} = \frac{0.75}

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(D) First, calculate the resistance of the galvanometer $(G)$: $G = \frac{V_g}{I_g} = \frac{0.75 \text{ V}}{10 \times 10^{-3} \text{ A}} = \frac{0.75}{0.01} = 75 \Omega$. To convert the galvanometer into an ammeter of range $I = 10 \text{ A}$, a shunt resistance $(S)$ is connected in parallel. The formula for shunt resistance is $S = \frac{I_g \times G}{I - I_g}$. Substituting the values: $S = \frac{10 \times 10^{-3} \times 75}{10 - 10 \times 10^{-3}} = \frac{0.75}{10 - 0.01} = \frac{0.75}{9.99}$. $S = \frac{75}{999} \Omega$.

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

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$A$ galvanometer has a current range of $10 \text{ mA}$ and a voltage range of $0.75 \text{ V}$. To convert this galvanometer into an ammeter of range $10 \text{ A}$, what is the shunt resistance?

MHT CET 2026 All MHT CET

A
$\frac{100}{999} \Omega$
B
$\frac{50}{999} \Omega$
C
$\frac{200}{999} \Omega$
D
$\frac{75}{999} \Omega$

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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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The deflection in a moving coil galvanometer is reduced to half when it is shunted with a $X \ \Omega$ coil. The relation between $X$ and the resistance of the galvanometer $G$ is:

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$A$ galvanometer has a current range of $10 \text{ mA}$ and a voltage range of $0.75 \text{ V}$. To convert this galvanometer into an ammeter of range $10 \text{ A}$, what is the shunt resistance?

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If a galvanometer is shunted by $(\frac{1}{n-1})^{th}$ of the value of its resistance, then the fraction of the total current passing through the galvanometer is

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