(C) Resistance of galvanometer,$G = 50 \Omega$. Full scale current,$i_g = 0.05 \text{ A}$. Maximum current to be measured,$i = 5 \text{ A}$. Area of c

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(C) Resistance of galvanometer,$G = 50 \Omega$. Full scale current,$i_g = 0.05 \text{ A}$. Maximum current to be measured,$i = 5 \text{ A}$. Area of cross-section,$A = 2.97 \times 10^{-2} \text{ cm}^2 = 2.97 \times 10^{-6} \text{ m}^2$. Specific resistance,$\rho = 5 \times 10^{-7} \Omega\text{-m}$. To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel. $S = \frac{i_g G}{i - i_g} = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \Omega$. Using the formula for resistance,$S = \rho \frac{l}{A}$,we get $l = \frac{S \cdot A}{\rho}$. $l = \frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}} = \frac{50}{99} \times \frac{29.7}{5} = 10 \times 0.3 = 3 \text{ m}$.

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

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$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

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Class 12 Questions

Class 12

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Chapter

Moving Charges and Magnetism

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

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Match the terms given in List-$I$ with suitable options provided in List-$II$ regarding the experiment to calculate the resistance and figure of merit of a galvanometer using the half-deflection method:

List-$I$ List-$II$

$A$. Figure of merit $(k)$ $P$. in series

$B$. Current sensitivity $(SI)$ $Q$. to get power loss minimized

$C$. Deflection is made half by using low resistance $R$. in shunt/parallel

$D$. High resistance box $S$. current per unit deflection

$T$. deflection per unit current

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(SI)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using low resistance	$R$. in shunt/parallel
$D$. High resistance box	$S$. current per unit deflection
	$T$. deflection per unit current

Medium

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$A$ current of $10^{-3} \ A$ is flowing through a resistance of $1000 \ \Omega$. To measure the correct potential difference across this resistance,a voltmeter must be used whose resistance should be:

Easy

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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:

Medium

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The sensitivity of a galvanometer is $60 \text{ division/A}$. When a shunt is used,its sensitivity becomes $10 \text{ division/A}$. If the galvanometer is of resistance $20 \ \Omega$,the value of shunt used is (in $Omega$)

DifficultTS EAMCET 2011

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The resistance of a galvanometer is $90\, \Omega$. If only $10\%$ of the main current is to flow through the galvanometer,how should a resistor be connected and what should be its value?

Medium

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$A$ current of $10^{-3} \ A$ is flowing through a resistance of $1000 \ \Omega$. To measure the correct potential difference across this resistance,a voltmeter must be used whose resistance should be:

$A$ 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:

The sensitivity of a galvanometer is $60 \text{ division/A}$. When a shunt is used,its sensitivity becomes $10 \text{ division/A}$. If the galvanometer is of resistance $20 \ \Omega$,the value of shunt used is (in $Omega$)

The resistance of a galvanometer is $90\, \Omega$. If only $10\%$ of the main current is to flow through the galvanometer,how should a resistor be connected and what should be its value?

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