Question

A moving coil galvanometer of resistance \[100{\text{ }}\Omega \] is used as an ammeter using a resistance \[{\text{0}}{\text{.1 }}\Omega \]. The maximum deflection current in the galvanometer is \[100{\text{ }}\mu {A_{}}\]. Find the minimum current in the circuit so that ammeter shows maximum deflection.
A. \[100.1{\text{ mA}}\]
B. \[{\text{1000}}{\text{.1 mA}}\]
C. \[{\text{10}}{\text{.01 mA}}\]
D. \[{\text{1}}{\text{.01 mA}}\]

Answer 1

Hint: To convert a galvanometer to an ammeter, we connect a shunt resistance parallel to the galvanometer resistance. Most of the current pass through the shunt resistance and as galvanometer and shunt resistance are connected in parallel; both have equal potential difference.

Formula used:
\[{I_g}{R_g} = \left( {I - {I_g}} \right)S\]
Where \[{I_g} = \] current passing through the galvanometer, \[{R_g} = \] resistance of the galvanometer, \[I = \] current passing through the whole circuit and \[S = \] shunt resistance.

Complete step by step answer:
Resistance of the galvanometer is \[{R_g} = 100{\text{ }}\Omega \].
Maximum deflection current in the galvanometer is, \[{I_g} = 100{\text{ }}\mu A\].
Shunt resistance connected in parallel, \[S = 0.1{\text{ }}\Omega \].

We need to find out the minimum current \[I\] in the circuit so that the ammeter shows maximum deflection.We know to convert a galvanometer to an ammeter; a shunt resistance is connected in parallel to the galvanometer resistance. As they are connected in parallel both, have equal potential differences, and the most amount of the current passes through shunt resistance.

The Potential difference across the galvanometer is \[{I_g}{R_g}\]. The Potential difference across the shunt resistance is \[\left( {I - {I_g}} \right)S\]. As they are connected in parallel,
\[{I_g}{R_g} = \left( {I - {I_g}} \right)S\]
Substituting the values we get,
\[ \Rightarrow {\text{100}} \times {\text{1}}{{\text{0}}^{ - 6}} \times 100 = \left( {I - 100 \times {{10}^{ - 6}}} \right) \times 0.1\]
\[ \Rightarrow 0.01 = \left( {I - 100 \times {{10}^{ - 6}}} \right) \times 0.1\]
Dividing both sides by \[0.1\]
\[ \Rightarrow 0.01 = \left( {I - 100 \times {{10}^{ - 6}}} \right) \times 0.1\]
\[ \Rightarrow I = 0.1 + {10^{ - 4}}\]
\[ \Rightarrow I = 0.1001{\text{ }}A\]
Therefore, the minimum current required in the circuit for maximum ammeter deflection is, \[\therefore I = 100.1{\text{ mA}}\]

Therefore, the correct option is A.

Note: In an ideal ammeter, the shunt resistance is negligible; thus, all the current passes through the shunt resistance. We get the actual current in the circuit whereas, in a real ammeter, there is some error in the ammeter reading as a small amount of current passes through the galvanometer. The smaller the value of shunt resistance the smaller is the error in ammeter reading.