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Ammeter - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Ammeter is considered one the most difficult concept.
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48 Questions around this concept.
Solve by difficulty
Consider a galvanometer shunted with 5 $\Omega$ resistance and 2% of current passes through it. What is the resistance of the given galvanometer?
A galvanometer having a coil resistance of 100 gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance (in $\Omega$ ) which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A , is :
Consider a circuit as shown in Fig. We want to measure the current flowing in the circuit.
For this, we connect an ammeter of resistance RA in the circuit. The percentage error in the current is
A galvanometer having coil resistance of 400 ohm gives a full-scale deflection of $2 m A$ is passed through it. The value of the resistance which can convert this galvanometer into an ammeter giving a full defection for a Current of $20 \mathrm{~A}$ is:
In an ammeter, 5% of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be:
A galvanometer of resistance R is shunted by a resistance S.The equivalent resistance of combination is
A shunt across a galvanometer
The ammeter in the fig. consists of a 500$\Omega$ coil connected in the parallel to 25$\Omega$ shunt. Find the reading of the ammeter.
A shunt across a galvanometer can be used to convert galvanometer to
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Ammeter
Ammeter
It is a device used to measure current and always connected in series with the circuit
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In the above circuit, A represents the ammeter.
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Conversion of galvanometer into ammeter: Connect a low resistance (shunt) in parallel to the galvanometer.
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The equivalent resistance of the combination is $\frac{G s}{G+s}$ where G is the galvanometer resistance and s is the shunt resistance.
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Required shunt $s=\frac{i_g G}{\left(i-i_g\right)}$. Where i is the total current, $i_g$ is the galvanometer current
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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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