AIPMT PRELIMS 2004 Physics Galvanometer MCQ Question

To convert a galvanometer into a voltmeter, we need to determine the series resistance required to allow the galvanometer to measure a specific voltage range without exceeding its deflection limit.

Given Data:

1. Resistance of galvanometer, $R_g = 50 \, \Omega$
2. Current for 1 division deflection, $I_g = 4 \times 10^{-4} \, \text{A}$
3. Total voltage range desired for the voltmeter, $V = 25 \, \text{V}$
4. Number of divisions, $N = 25$

Step 1: Find the full-scale current for 25 divisions

The full-scale current, $I$, that corresponds to the maximum voltage (25 V) can be calculated using Ohm's law:

$$I = \frac{V}{R_g}$$

However, we first need to calculate the current that corresponds to the full-scale deflection of the galvanometer, which occurs when it shows 25 divisions.

The total current through the galvanometer for 25 divisions is given by:

$$I = N \cdot I_g = 25 \cdot 4 \times 10^{-4} \, \text{A} = 1 \times 10^{-2} \, \text{A} = 0.01 \, \text{A}$$

Step 2: Calculate the total resistance required for the voltmeter

To find the total resistance $R_{total}$ required for this current at 25 V, we use Ohm's law again:

$$R_{total} = \frac{V}{I} = \frac{25 \, \text{V}}{0.01 \, \text{A}} = 2500 \, \Omega$$

Step 3: Calculate the resistance needed in series

The total resistance in the circuit will be the sum of the galvanometer's resistance and the series resistance $R_s$:

$$R_{total} = R_g + R_s$$

Substituting the values we have:

$$2500 \, \Omega = 50 \, \Omega + R_s$$

Now, solving for $R_s$:

$$R_s = 2500 \, \Omega - 50 \, \Omega = 2450 \, \Omega$$

Conclusion

Thus, to convert the galvanometer into a voltmeter with a range of 25 V, we need to connect a resistance of $2450 \, \Omega$ in series.

Explanation of Options

• A) 245 Ω as a shunt: Incorrect. A shunt would not increase the resistance adequately or function as a series resistor for voltage measurement.
• B) 2550 Ω in series: Incorrect. This value exceeds the required resistance.
• C) 2450 Ω in series: Correct. This matches our calculated resistance needed to achieve the desired voltage range.
• D) 2500 Ω as a shunt: Incorrect. A shunt would reduce the current passing through the galvanometer rather than adding to the overall series resistance.

Therefore, the correct answer is C) 2450 Ω in series.