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Step-by-Step Solution
Step 1: Write down the initial condition for the 0–1 V range
The galvanometer has resistance $G$ and is converted into a voltmeter of range 0–1 V by connecting a resistor $R_1$ in series. Let $i_g$ be the maximum current through the galvanometer when the voltmeter reads full scale (1 V). Thus,
$1 = i_g \bigl(G + R_1\bigr)\,.$
This equation means that when a current $i_g$ flows through the series combination of $G + R_1$, the potential difference across it is 1 V.
Step 2: Write down the condition for the 0–2 V range
To increase the voltmeter range from 1 V to 2 V, we connect an additional resistor $R_2$ in series with $R_1$. With the same galvanometer current $i_g$ at the new full-scale deflection (2 V range), we have:
$2 = i_g \bigl(G + R_1 + R_2\bigr)\,.$
Step 3: Form the ratio to find $R_2$
Divide the first equation by the second to eliminate $i_g$:
$ \frac{1}{2} = \frac{i_g (G + R_1)}{i_g (G + R_1 + R_2)} \,.$
Simplifying,
$ \frac{1}{2} = \frac{G + R_1}{G + R_1 + R_2}\,.$
Step 4: Solve for $R_2$
Cross multiply:
$G + R_1 + R_2 = 2(G + R_1)\,. $
This simplifies to:
$R_2 = (G + R_1) \,.$
Step 5: State the final answer
Therefore, the additional resistance that must be connected in series with $R_1$ to increase the voltmeter range from 0–1 V to 0–2 V is
$R_2 = R_1 + G.$
