Find the resistance that must be placed in parallel with a \(25.0 - \Omega \) galvanometer having a \(50.0 - \mu A\) sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a \(300 - mA\) full-scale reading.

A galvanometer is an electromechanical device used to detect electric current. A galvanometer deflects a pointer in response to an electric current flowing through a coil in a constant magnetic field. An example of an actuator is a galvanometer.

The total flow of electrons via a wire can be used to describe the rate of electron flow. Anything that prevents current flow is referred to as "resistance." An electrical circuit needs resistance in order to transform electrical energy into light, heat, or movement.

• Inner resistance in galvanometer: \(25.0 - \Omega \)
• Sensitivity of galvanometer:\(50.0 - \mu A\)
• Scale measure of ammeter: \(300 - mA\)

Calculate the additional resistance \(R\), connected in parallel to the internal resistance, necessary for the galvanometer tohave a \({I_{tot}} = 300{\rm{ }}mA\) full-scale reading. Since the resistances are connected in parallel, they have the same potentialdifference. When \({I_{tot}}\) goes through the ammeter, part of it \(\left( {{I_g}} \right)\) goes through the galvanometer, and part \(\left( {{I_R}} \right)\) goesthrough the additional resistance, which means that we have \({I_{tot}} = {I_g} + {I_R}\). It is needed \({I_g}\), to be maximal (equal to thesensitivity) when the external current is equal to \({I_{tot}} = 300{\rm{ }}mA\). The resistance \(R\) is then –

\(\begin{align}{c}{I_g}r &= {I_R}R\\R &= \frac{{{I_g}r}}{{{I_R}}}\\ &= \frac{{{I_g}r}}{{{I_{tot}} - {I_g}}}\end{align}\)

\(\begin{align}{} &= \frac{{(50{\rm{ }}\mu A) \cdot (25{\rm{ }}\Omega )}}{{\left( {0.3 - 50 \cdot {{10}^{ - 6}}} \right)A}}\\ &= 4.17\;m\Omega \end{align}\)

Therefore, the value for resistance is obtained as \(R = 4.17\;m\Omega \).