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Ammeter is considered one the most difficult concept.
28 Questions around this concept.
A miliammeter of range 10 mA has a coil of resistance $1 \Omega$. To use it as an ammetre of range 1 A , the required shunt must have a resistance of
A galvanometer has a resistance of $50 \Omega$ and its full scale deflection current is $50 \mu A$. What shunt resistance should be added so that the ammeter can have a range of $0-5 \mathrm{~mA}$ ?
Two resistors 400 ohm and 800 ohm are connected in series with a 6V battery. It is desired to measure the current in the circuit. An ammeter of 10 ohm resistance is used for this purpose. The reading in the ammeter is:
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The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times. The shunt used is :
The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times.
The shunt used is :
Find the reading of the ammeters (in ampere) connected as shown in the network.
In the circuit as shown in figure, the current in ammeter is:
A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading up to 0.06 A when a shunt of resistance r is connected across it. What is the maximum current which can be sent through this galvanometer, if no shunt is used?
A galvanometer having 50 divisions provided with a variable shunt which is used to measure the current when connected in series with a resistance of and a battery of internal resistance .It is observed that when the shunt resistances are and ,the deflection are respectively 9 and 30 division. The resistance of the Galvanometer is :
The reading of ammeter in the figure shown below is
Ammeter
It is a device used to measure current and always connected in series with the circuit
In the above circuit, A represents the ammeter.
Conversion of galvanometer into ammeter: Connect a low resistance (shunt) in parallel to the galvanometer.
The equivalent resistance of the combination is $\frac{G s}{G+s}$ where G is the galvanometer resistance and s is the shunt resistance.
Required shunt $s=\frac{i_g G}{\left(i-i_g\right)}$. Where i is the total current, $i_g$ is the galvanometer current
To pass $n^{\text {th }}$ part of main current through galvanometer that is if $i_g$ $=i / n$
Then the required shunt resistance $s=\frac{G}{(n-1)}$
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