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Step 1: Identify the Known Quantities
• Galvanometer internal resistance: 25\,\Omega
• External series resistance (initially): 3000\,\Omega
• Battery voltage: 2\,\mathrm{V}
• Initial full-scale deflection: 30 units
• Desired deflection: 10 units
• Additional series resistance to be found: R
Step 2: Relate Deflection to Current
The deflection D of the galvanometer is directly proportional to the current I passing through it. We can write:
D = k \, I
where k is the proportionality constant.
Step 3: Write Current Expressions for Both Deflections
1. When the deflection is 30 units, the total resistance in the circuit is (25 + 3000) = 3025\,\Omega. The current is:
I_{\text{initial}} = \frac{2}{3025}.
According to the proportionality,
30 = k \cdot \frac{2}{3025}.
2. When an additional resistance R is added to reduce the deflection to 10 units, the total resistance becomes (25 + 3000 + R) = (3025 + R)\,\Omega. The current then is:
I_{\text{reduced}} = \frac{2}{3025 + R}.
Hence,
10 = k \cdot \frac{2}{3025 + R}.
Step 4: Form a Ratio to Eliminate k
Divide the equation for 30 units deflection by the equation for 10 units deflection:
\frac{30}{10}
=
\frac{\frac{2}{3025}}{\frac{2}{3025 + R}}.
Simplifying,
3 = \frac{3025 + R}{3025}.
Step 5: Solve for R
From
3 = \frac{3025 + R}{3025},
multiply both sides by 3025:
3025 + R = 3 \times 3025 = 9075.
Therefore,
R = 9075 - 3025 = 6050\,\Omega.
Final Answer
The additional resistance required is 6050\,\Omega.
