1. Given data

• Galvanometer resistance: $G = 100\,\Omega$
• Number of divisions: $50$
• Sensitivity: $20\,\mu A$/division

So full-scale current of the galvanometer is $I_g = 50 \times 20\,\mu A = 1000\,\mu A = 1\,mA$

2. Voltmeter conversion formula

To convert a galvanometer into a voltmeter of range $V$, a resistance $R$ is connected in series such that $V = I_g(G+R)$ Hence, $R = \frac{V}{I_g} - G$

Since $I_g = 1\,mA = 10^{-3}\,A$,

For range $0-2\,V$

$R_1 = \frac{2}{10^{-3}} - 100 = 2000 - 100 = 1900\,\Omega$

For range $0-10\,V$

$R_2 = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\,\Omega$

For range $0-20\,V$

$R_3 = \frac{20}{10^{-3}} - 100 = 20000 - 100 = 19900\,\Omega$

3. How multi-range voltmeter is made

For a multi-range voltmeter, resistors are arranged in series with tapping points so that:

• for $2\,V$: total series resistance with galvanometer should be $1900\,\Omega$
• for $10\,V$: total series resistance with galvanometer should be $9900\,\Omega$
• for $20\,V$: total series resistance with galvanometer should be $19900\,\Omega$

Let the added resistors be $r_1, r_2, r_3$ in series. Then, $r_1 = 1900\,\Omega$ $r_1 + r_2 = 9900\,\Omega \Rightarrow r_2 = 8000\,\Omega$ $r_1 + r_2 + r_3 = 19900\,\Omega \Rightarrow r_3 = 10000\,\Omega$

Thus the correct circuit must have three series resistors: $1900\,\Omega,\; 8000\,\Omega,\; 10000\,\Omega$ with proper tapping points to select ranges $2\,V$, $10\,V$, and $20\,V$.

4. Matching with options

The option showing a galvanometer in series with resistors $1900\,\Omega$, $8000\,\Omega$, and $10000\,\Omega$ with suitable selector/tapping arrangement is Option C.

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.