Question

A galvanometer has a resistance of 100 ohm. A shunt resistance of 1 ohm is connected across it. What part of total current flows through the galvanometer?

Answer 1

Use the formula for the shunt resistance that is to be connected in parallel with the galvanometer. This formula gives the relation between the current for which the galvanometer shows full scale deflection, resistance of the galvanometer and maximum range of the total current in the circuit measured by the ammeter.

Formula used:
The value of the resistance $S$ of the shunt resistor that should be connected in the circuit to convert the galvanometer into ammeter is given by
$S = \dfrac{{{I_g}G}}{{I - {I_g}}}$ …… (1)
Here, ${I_g}$ is the current for which the galvanometer shows full scale deflection, $G$ is the resistance of the galvanometer and $I$ is the total current in the circuit.

Complete step by step answer:

We have given that the resistance of the galvanometer is $100\,\Omega$.
$G = 100\,\Omega$
We have also given that the resistance of the shunt resistor is $1\,\Omega$.
$S = 1\,\Omega$
We have asked to determine the part of the total current that flows through the galvanometer.
Substitute $100\,\Omega$ for $G$ and $1\,\Omega$ for $S$ in equation (1).
$1\,\Omega = \dfrac{{{I_g}\left( {100\,\Omega } \right)}}{{I - {I_g}}}$
$\Rightarrow 1 = \dfrac{{100{I_g}}}{{I - {I_g}}}$
$\Rightarrow I - {I_g} = 100{I_g}$
$\Rightarrow I = 100{I_g} + {I_g}$
$\Rightarrow I = 101{I_g}$
$\Rightarrow \dfrac{{{I_g}}}{I} = \dfrac{1}{{101}}$

Hence, the part of the total current that flows through the galvanometer is $\dfrac{1}{{101}}$.

Note: The students should keep in mind that we have asked to determine the part of the total current (which is the maximum range of the ammeter in the circuit) that flows into the galvanometer (which is the current for which the galvanometer shows full scale deflection) which the ratio of current through galvanometer to total current and not the ratio of total current to current through the galvanometer.