A moving coil galvanometer has a resistance of $ 10\, O $ and full scale deflection of $ 0.01\, A $ . It can be converted into voltmeter of $ 10\, V $ full scale by connecting into resistance of :
Let $G$ be resistance of galvanometer and $i_g$ the current through it. Let $V$ is maximum potential difference, then from Ohm?? law
$i_{g} = \frac{V}{G + R}$ $\Rightarrow \,R = \frac{V}{i_{g}} - G$ Given, $G = 10\,\Omega, i_{g} = 0.01\,A$ $V = 10$ volt $\therefore \, R=\frac{10}{0.01}-10=990\, \Omega$ Thus, on connecting a resistance $R$ of $990 \Omega$ in series with the galvanometer, the galvanometer will become a voltmeter of range zero to $10\, V$. For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.
There are various electrical instruments used to measure current, power, voltage, etc. Some of them are briefly explained below:
Moving Coil Galvanometer
It is an electromagnetic device which measures small values of current.
Its working principle is that whenever a current loop is placed in a magnetic field, it experiences a certain torque. The value of that torque can be modified by modifying the current in the loop.
For a current carrying loop having N turns, and cross sectional area A, carrying current i, whenever it is placed in and along the direction of an external magnetic field B, it experiences a torque given by:
