Question

What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{‐}\mathbf{k}\mathbf{}\mathbf{\Omega }$resistance on its 100-Vscale?

Short answer

The sensitivity of the galvanometer inside a voltmeter that has a $25.0\mathrm{k\Omega }$resistance on its 100 V scale is, $\mathrm{I}=4.0\mathrm{mA}$.

Step-by-step solution

Concept Introduction

A galvanometer is an electric current measurement instrument that is electromechanical. A galvanometer responds to an electric current running through a coil in a constant magnetic field by deflecting a pointer. Galvanometers can be viewed as a type of actuator.

The rate of electron flow can be described as the aggregate flow of electrons via a wire. The term "resistance" refers to anything that stands in the way of current flow. To convert electrical energy to light, heat, or movement, an electrical circuit must have resistance.

Information Provided

• Resistance in galvanometer:$25.0\mathrm{k\Omega }\left(\frac{{10}^3\mathrm{\Omega }}{1\mathrm{k\Omega }}\right)=2.50\times {10}^4\mathrm{\Omega }$
• Scale measure of galvanometer: 100 V

Calculation for Current

100 V is the maximum potential the galvanometer can read. This means that its sensitivity to currents can be calculated by calculating what amount of current causes a potential difference of 30 V .

Mathematically it can be written as,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ =\frac{100\mathrm{V}}{2.50\times {10}^4\mathrm{\Omega }} \\ =4.0\times {10}^{-3}\mathrm{A}\left(\frac{1\mathrm{mA}}{{10}^{-3}\mathrm{A}}\right) \\ =4.0\mathrm{mA} \end{gathered}$$

Therefore, the value for current is obtained as $\mathrm{I}=4.0\mathrm{mA}$.