Question

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt the resistance in ohm needed to be connected in series with the coil will be
(A) 1000
(B) 100000
(C) 99995
(D) 9995

Answer 1

Hint If each division of the galvanometer were reading 1 volts, then the maximum volt readable by the galvanometer (full scale deflection) would be 150 V. Use the voltage and current sensitivity of the galvanometer to find the resistance of the galvanometer itself.
Formula used: In this solution we will be using the following formulae;
$R = \dfrac{V}{I}$ where $R$ is the resistance of a component, $V$ is the potential difference across the component, and $I$ is the current flowing through the component.
${R_{eqs}} = {R_1} + {R_2}$ where ${R_{eqs}}$ is the equivalent resistance of two resistances in series, and ${R_1},{R_2}$ are the individual resistances.

Complete Step-by-Step solution:
Now, the moving galvanometer has 150 equal divisions with 10 divisions per milli-ampere. Hence, the maximum current that can be measured by the galvanometer this way is
${I_g} = \dfrac{{150div}}{{10div/mA}} = 15mA$
And similarly, the maximum voltage which can be measured by the galvanometer is
${V_g} = \dfrac{{150div}}{{2div/mA}} = 75mV$.
With this maximum, we can calculate the resistance of the galvanometer itself using the formula
$R = \dfrac{V}{I}$ where $R$ is the resistance of a component, $V$ is the potential difference across the component, and $I$ is the current flowing through the component.
Hence,
$R = \dfrac{{{V_g}}}{{{I_g}}} = \dfrac{{75mV}}{{15mA}} = 5\Omega$
Now, we want the galvanometer to read 1 volt per division, which means that the maximum should be
${V_{g2}} = 150div \times 1\dfrac{V}{{div}} = 150V$
Hence, the resistance of the galvanometer should be
${R_{g2}} = \dfrac{{150A}}{{0.015A}} = 10000\Omega$
Hence, since the resistance added in series should be
${R_s} = 10000 - 5 = 9995\Omega$

Hence, the correct option is D

Note: For clarity, we subtract to get the resistance above because, 1000 ohms is the resistance the galvanometer should have but the galvanometer has 5 ohms. Hence, since the series equivalent resistance is given by
${R_{eqs}} = {R_1} + {R_2}$ where ${R_{eqs}}$ is the equivalent resistance of two resistances in series, and ${R_1},{R_2}$ are the individual resistances.
Hence,
$10000\Omega = {R_s} + 5\Omega$
$\Rightarrow {R_s} = 10000 - 5 = 9995\Omega$