(A) Given:Resistance of galvanometer, $G = 100 \Omega$Shunt resistance, $S = 0.1 \Omega$Maximum deflection current in galvanometer, $I_g = 100 \mu A =

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(A) Given:Resistance of galvanometer, $G = 100 \Omega$Shunt resistance, $S = 0.1 \Omega$Maximum deflection current in galvanometer, $I_g = 100 \mu A = 100 \times 10^{-6} A = 10^{-4} A$For an ammeter, the galvanometer and shunt resistance are connected in parallel. The potential difference across both is the same:$V_{AB} = I_g G = (I - I_g) S$Substituting the values:$(100 \times 10^{-6}) \times 100 = (I - 100 \times 10^{-6}) \times 0.1$$10^{-2} = (I - 10^{-4}) \times 10^{-1}$$0.1 = I - 0.0001$$I = 0.1 + 0.0001 = 0.1001 A$Converting to milliamperes:$I = 0.1001 \times 1000 mA = 100.1 mA$Thus, the total current in the circuit is $100.1 mA$.

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

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$A$ moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a shunt resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu A$. Find the total current in the circuit so that the ammeter shows maximum deflection. (in $mA$)

TS EAMCET 2020 All TS EAMCET

A
$100.1$
B
$1000.1$
C
$10.01$
D
$1.01$

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Class 12

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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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When a shunt resistance of $4r$ is connected to a galvanometer,it becomes an ammeter that can measure $0.03 \, A$. When a shunt resistance of $r$ is connected to the same galvanometer,it becomes an ammeter that can measure $0.06 \, A$. What is the current capacity $(i_g)$ of the galvanometer in $A$?

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In order to increase the sensitivity of a moving coil galvanometer,one should decrease

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The resistance of a galvanometer is $50\,\Omega$ and it requires $2\,\mu A$ per two division deflection. The value of the shunt required in order to convert this galvanometer into an ammeter of range $5\,A$ is (The number of divisions on the galvanometer scale on one side is $30$).

Medium

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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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$A$ galvanometer of resistance $G$ is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged,the resistance to be put in series with the same galvanometer is

MediumMHT CET 2022

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$A$ moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a shunt resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu A$. Find the total current in the circuit so that the ammeter shows maximum deflection. (in $mA$)

Similar Questions

In order to increase the sensitivity of a moving coil galvanometer,one should decrease

The resistance of a galvanometer is $50\,\Omega$ and it requires $2\,\mu A$ per two division deflection. The value of the shunt required in order to convert this galvanometer into an ammeter of range $5\,A$ is (The number of divisions on the galvanometer scale on one side is $30$).

$A$ galvanometer of resistance $G$ is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged,the resistance to be put in series with the same galvanometer is

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