Question

A galvanometer of resistance $G$ is converted into a voltmeter of range $(0 - V)$ by connecting a resistor of $250\ \Omega$ with it. If resistor of $250\ \Omega$ is replaced by another resistor of $900\ \Omega$, its range becomes $(0 - 3\text{ V})$. The resistance $G$ of the galvanometer is :

• A. $150\ \Omega$
• B. $125\ \Omega$
• C. $100\ \Omega$
• D. $75\ \Omega$

Hint:

When the range of a voltmeter is scaled by a factor of $n$ (i.e., $V' = nV$), the relation between the multipliers is $R' - nR = (n - 1)G$. For $n = 3$, $G = \frac{R_2 - 3R_1}{2} = \frac{900 - 750}{2} = 75\ \Omega$.\

Answer 1

The Correct Option is D

Concept:
• A galvanometer is converted into a voltmeter of a desired voltage range by connecting a high resistance in series with the galvanometer coil.
• The total potential difference measured by the voltmeter is the product of full-scale deflection current $I_g$ and the total series resistance $(G + R)$.
• The fundamental relation governing the voltmeter conversion is given by $V = I_g(G + R)$.

Step 1: Formulate the equation for the first case
In the initial configuration, the series resistor is $R_1 = 250\ \Omega$ and the maximum measurable voltage range is $V_1 = V$.
Let $I_g$ be the current required for full-scale deflection of the galvanometer.
Applying Ohm's law across the series combination:
\[ V = I_g(G + 250) \quad \text{--- (Equation 1)} \]

Step 2: Formulate the equation for the second case
In the modified configuration, the series resistor is replaced by $R_2 = 900\ \Omega$, and the voltage range increases to $V_2 = 3V$.
Applying the same relation for the new range:
\[ 3V = I_g(G + 900) \quad \text{--- (Equation 2)} \]

Step 3: Solve the two equations simultaneously
Divide Equation 2 by Equation 1 to eliminate the full-scale current $I_g$ and voltage $V$:
\[ \frac{3V}{V} = \frac{I_g(G + 900)}{I_g(G + 250)} \]
Simplify the ratio:
\[ 3 = \frac{G + 900}{G + 250} \]
Cross-multiply to solve for the unknown galvanometer resistance $G$:
\[ 3(G + 250) = G + 900 \]
\[ 3G + 750 = G + 900 \]
Rearrange the terms containing $G$ to one side:
\[ 3G - G = 900 - 750 \]
\[ 2G = 150 \]
\[ G = \frac{150}{2} = 75\ \Omega \]

Step 4: Conclusion
The internal resistance of the galvanometer coil is $75\ \Omega$, which matches option (D).