
The ratio of a shunt resistance and the resistance of a galvanometer is $1:499$. If the full scale deflection current of the galvanometer is $2mA$, the range of the ammeter is:
A. $2A$
B. $1.4A$
C. $1.6A$
D. $1A$
Answer
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Hint: We know that shunt is used in the galvanometer for the measurement of larger currents. The direction of flow of current in any circuit is determined by the pointer of the galvanometer. A galvanometer can be converted into ammeter by connecting low resistance (shunt) in parallel to the galvanometer. If the resistance is connected to a series with a galvanometer then it is used as a voltmeter.
Complete step-by-step answer:
A galvanometer is an ammeter but all ammeters are not galvanometers. It can read a smaller current with greater accuracy. Let the shunt of ${R_s}$ is connected in parallel to galvanometer of resistance \[{R_g}\] and \[{i_g},{i_s}\] are the current flowing in galvanometer and shunt resistance respectively, We can draw the circuit of galvanometer as shown in figure ;
$(V = IR)$
We know that for the parallel connection of galvanometer resistance and shunt their voltage should be the same. We know that by ohm’s law, hence;
$
{i_g} \times {R_g} = {i_s} \times {R_s} \\
\dfrac{{{R_s}}}{{{R_g}}} = \dfrac{1}{{499}} \\
{i_s} = {i_g}\dfrac{{499}}{1} = 2 \times 499 = 998mA \\
$
Now applying Kirchhoff’s current law we will get the full scale deflection current of the galvanometer,
$i = {i_g} + {i_s} = 2 + 998 = 1000mA = 1A$
Therefore option D is the correct option for these questions hence The ratio of a shunt resistance and the resistance of a galvanometer is $1:499$. If the full scale deflection current of the galvanometer is $2mA$ , the range of the ammeter is $1A$.
So, the correct answer is “Option D”.
Note: The shunt resistor is made up of the material which has a low temperature coefficient of resistance. It is connected parallel with the ammeter so that the range of the galvanometer can be extended. In this problem we have calculated the full scale deflection current with the help of a circuit diagram of the galvanometer.
Complete step-by-step answer:
A galvanometer is an ammeter but all ammeters are not galvanometers. It can read a smaller current with greater accuracy. Let the shunt of ${R_s}$ is connected in parallel to galvanometer of resistance \[{R_g}\] and \[{i_g},{i_s}\] are the current flowing in galvanometer and shunt resistance respectively, We can draw the circuit of galvanometer as shown in figure ;
$(V = IR)$
We know that for the parallel connection of galvanometer resistance and shunt their voltage should be the same. We know that by ohm’s law, hence;
$
{i_g} \times {R_g} = {i_s} \times {R_s} \\
\dfrac{{{R_s}}}{{{R_g}}} = \dfrac{1}{{499}} \\
{i_s} = {i_g}\dfrac{{499}}{1} = 2 \times 499 = 998mA \\
$
Now applying Kirchhoff’s current law we will get the full scale deflection current of the galvanometer,
$i = {i_g} + {i_s} = 2 + 998 = 1000mA = 1A$
Therefore option D is the correct option for these questions hence The ratio of a shunt resistance and the resistance of a galvanometer is $1:499$. If the full scale deflection current of the galvanometer is $2mA$ , the range of the ammeter is $1A$.
So, the correct answer is “Option D”.
Note: The shunt resistor is made up of the material which has a low temperature coefficient of resistance. It is connected parallel with the ammeter so that the range of the galvanometer can be extended. In this problem we have calculated the full scale deflection current with the help of a circuit diagram of the galvanometer.
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