(A) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer coil.The formula for the series

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$3950 \Omega$ in series with galvanometer

(A) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer coil.The formula for the series resistance is given by $R = \frac{V}{I_g} - G$.Given values are:Voltage range $V = 20 \text{ V}$Full scale deflection current $I_g = 5 \text{ mA} = 5 \times 10^{-3} \text{ A}$Galvanometer resistance $G = 50 \Omega$Substituting these values into the formula:$R = \frac{20}{5 \times 10^{-3}} - 50$$R = \frac{20}{0.005} - 50$$R = 4000 - 50 = 3950 \Omega$Thus,a resistance of $3950 \Omega$ must be connected in series with the galvanometer.

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

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$A$ galvanometer coil has a resistance of $50 \Omega$ and the meter shows full scale deflection for a current of $5 \text{ mA}$. This galvanometer is converted into a voltmeter of range $0-20 \text{ V}$ by connecting:

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$3950 \Omega$ in series with galvanometer
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$4050 \Omega$ in series with galvanometer
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$3950 \Omega$ in parallel with galvanometer
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$4050 \Omega$ in parallel with galvanometer

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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 coil has a resistance $90\, \Omega$ and full-scale deflection current $10\, mA$. $A$ $910\, \Omega$ resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.1\, V$,the number of divisions on its scale is:

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$A$ $50\, \Omega$ galvanometer gets full-scale deflection when a current of $0.01\, A$ passes through the coil. When it is converted to a $10\, A$ ammeter,the shunt resistance is ........... $\Omega$.

Medium

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$A$ voltmeter has a range $0-V$ with a series resistance $R$. With a series resistance $2R$,the range is $0-V'$. The correct relation between $V$ and $V'$ is

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$A$ student connects a voltmeter,an ammeter,and a resistor as shown in the circuit. If the voltmeter reading is $20 \ V$ and the ammeter reading is $4 \ A$,then the resistance $R$ will be:

Medium

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If only $\frac{1}{51}$ of the main current is to be passed through a galvanometer,then the shunt required is $R_1$. If only $\frac{1}{11}$ of the main voltage is to be developed across the galvanometer,then the resistance required is $R_2$. Find the ratio $\frac{R_2}{R_1}$.

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$A$ galvanometer coil has a resistance of $50 \Omega$ and the meter shows full scale deflection for a current of $5 \text{ mA}$. This galvanometer is converted into a voltmeter of range $0-20 \text{ V}$ by connecting:

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$A$ galvanometer coil has a resistance $90\, \Omega$ and full-scale deflection current $10\, mA$. $A$ $910\, \Omega$ resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.1\, V$,the number of divisions on its scale is:

$A$ $50\, \Omega$ galvanometer gets full-scale deflection when a current of $0.01\, A$ passes through the coil. When it is converted to a $10\, A$ ammeter,the shunt resistance is ........... $\Omega$.

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