Physics > Current Electricity > 10.0 Electrical devices
Current Electricity
1.0 Introduction
2.0 Conduction of current in a metal
3.0 Ohm's law
3.1 Temperature dependence of resistance
3.2 Resistivities of different materials
3.3 Limitations of ohm's law
4.0 Combination of Resistors
5.0 Electromotive force $\left( \xi \right)$
6.0 Heating effect of current
7.0 Wheatstone bridge
8.0 Metre Bridge Or Slide wire bridge
9.0 Potentiometer
9.1 Comparison of emfs of two primary cells.
9.2 Determination of Internal resistance of a cell using potentiometer
10.0 Electrical devices
10.2 Voltmeter
3.2 Resistivities of different materials
3.3 Limitations of ohm's law
9.2 Determination of Internal resistance of a cell using potentiometer
It is an instrument which is used to measure potential difference across two points in a circuit.
It connected in parallel along the points for which the potential difference is to be measured.
An ideal voltmeter has infinite resistance.
A galvanometer can be converted into a voltmeter by connecting a very high resistance in series with it.
The connected resistance should be very large as compared to the resistance of any circuit element with which the voltmeter is connected.
Mathematically, $$\begin{equation} \begin{aligned} \Rightarrow {I_g}({R_v} + G) = V \\ \Rightarrow {R_v} = \frac{V}{{{I_g}}} - G \\\end{aligned} \end{equation} $$
Question 33. A galvanometer having a coil resistance of $100\Omega $ gives a full scale deflection when a current of 1 $mA$ is passed through it. What is the value of the resistance which can convert this galvanometer into an ammeter giving full scale deflection for a current of 10 $A$?
Solution:
Given: $$\begin{equation} \begin{aligned} {R_g} = 100\Omega \\ {I_g} = 1mA = 0.001A \\ I = 10A \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} \Rightarrow S = \frac{{{I_g}G}}{{(I - {I_g})}} \\ \Rightarrow S = \frac{{(0.001)(100)}}{{(10 - 0.001)}} \\ \Rightarrow S = 0.01\Omega \\\end{aligned} \end{equation} $$
Question 34.A galvanometer having a coil resistance of $100\Omega $ gives a full-scale deflection when a current of 1 $mA$ is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full-scale deflection for a potential difference of 10 $V$?
Solution:
Given: $$\begin{equation} \begin{aligned} {R_g} = 100\Omega \\ {I_g} = 1mA = 0.001A \\ V = 10V \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} \Rightarrow {R_v} = \frac{V}{{{I_g}}} - G \\ \Rightarrow {R_v} = \frac{{10}}{{0.001}} - 100 \\ \Rightarrow {R_v} = 10000 - 100 \\ \Rightarrow {R_v} = 9900\Omega \\\end{aligned} \end{equation} $$
