Question

A galvanometer of $50\Omega $ resistance has $25$ divisions. A current of $4 \times {10^{ - 4}}A$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25V$,it should be connected to resistance of
A. $2450\Omega $ in series
B. $2500\Omega $ in series
C. $245\Omega $ in series
D. $2550\Omega $ in series

Answer 1

Hint:In this question, we are having a galvanometer with given resistance and to convert that galvanometer to a voltmeter we have to replace it with a resistance. To find that value we are using the famous law known as ohm’s law. By substituting the given values, we can find our answer.

Complete Step by Step Answer:
Firstly, writing the given values from the question:
Resistance of galvanometer ${R_g} = 50\Omega $
Range of voltage $ = 25V$
Current of the galvanometer $\left( {{I_g}} \right)$ be given by product of current and the divisions of the galvanometer.
${I_g} = \left( {4 \times {{10}^{ - 4}}} \right) \times 25$
$\Rightarrow {I_g} = {10^{ - 2}}A$
In this question we had a series circuit having a galvanometer and to convert it into a voltmeter, we have to connect a resistance which should have value. Hence, using Ohm’s law
$V = {I_g}\left( {R + {R_g}} \right)$
Substituting the above values
$25 = {10^{ - 2}}\left( {R + 50} \right)$
Solving the mathematical expression
$
R = 25 \times {10^2} - 50 \\
\Rightarrow R = 2500 - 50 \\
\therefore R = 2450\Omega \\ $
Hence, we solved our problem. The correct option is A.

Note:Galvanometer is a device used to measure the electrical current along the circuit. voltmeter is a device used to measure the electrical potential difference between two points in the electric circuit. Besides we all know about the ammeter which is also used to measure the current then what’s the difference between the galvanometer then the answer is galvanometer measures the magnitude as well as direction of current along the circuit.