Question:

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 :

Updated On: Jun 20, 2022
  • $ 9.90 \,O $ in series
  • $ 10 \,O $ in series
  • $ 990 \,O $ in series
  • $ 0.10 \,O $ in series
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The Correct Option is C

Solution and Explanation

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.
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Concepts Used:

Electrical Instruments

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:

ԏ = NiAB

moving coil galvanometer