A galvanometer has 30 divisions and a sensitivity $16\ \mu ext{A/div}$. It can be converted into a voltmeter

Question

A galvanometer has 30 divisions and a sensitivity $16\ \mu	ext{A/div}$. It can be converted into a voltmeter to read $3\ 	ext{V}$ by connecting (approximately):

Accepted Answer

1.  **Calculate Full Scale Current ($I_g$):** The galvanometer has 30 divisions and a current sensitivity (current per division) of $16\ \mu	ext{A/div}$.
    $I_g = 30 	imes 16\ \mu	ext{A} = 480\ \mu	ext{A} = 480 	imes 10^{-6}\ 	ext{A}$ .
2.  **Principle of Conversion:** To convert a galvanometer into a voltmeter, a high resistance ($R$) must be connected in **series** with the galvanometer coil . This ensures the galvanometer draws a very small current while measuring the potential difference.
3.  **Calculation:** The total resistance ($R_{eq}$) required for the circuit to measure a voltage $V$ with a full-scale current $I_g$ is given by Ohm's Law:
    $R_{eq} = R + R_g = \frac{V}{I_g}$
    where $R_g$ is the internal resistance of the galvanometer.
    $R + R_g = \frac{3}{480 	imes 10^{-6}} = \frac{3000000}{480} = 6250\ \Omega = 6.25\ 	ext{k}\Omega$.
4.  **Approximation:** The added series resistance $R = 6250\ \Omega - R_g$. Since the galvanometer resistance $R_g$ is typically small (order of $10-100\ \Omega$), the required series resistance is approximately $6.25\ 	ext{k}\Omega$. The closest option is nearly $6\ 	ext{k}\Omega$ in series.

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