Question

A galvanometer of resistance 50 Ω is connected to the battery of 3 V along with a resistance 2950 Ω in series. A full scale deflections of 30 divisions is obtained in galvanometer. In order to reduce this deflections to 20 division, the resistance in series should be

• A. 5050 Ω
• B. 6050 Ω
• C. 4450 Ω
• D. 5550 Ω

Answer 1

The Correct Option is C

Let's calculate the current for the full-scale deflection: 

\( I_1 = \frac{V}{R_{\text{gal}} + R_{\text{series}}} \)

\( I_1 = \frac{3V}{50\Omega + 2950\Omega} \)

\( I_1 = \frac{3V}{3000\Omega} \)

\( I_1 = 0.001 \, \text{A} \)

Now, let's calculate the current required for a 20-division deflection:

\( I_2 = \frac{2}{3} \cdot I_1 \)

\( I_2 = \frac{2}{3} \times 0.001 \, \text{A} \)

\( I_2 = 0.000667 \, \text{A} \)

To find the resistance needed, we can use Ohm's law:

\( V = I \cdot R \)

For the 20-division deflection:

\( 0.000667 \, \text{A} \times (50 \, \Omega + R) = 3 \, \text{V} \)

Solving this equation for \( R \), we get:

\( 0.000667 \, \text{A} \times R = 3 \, \text{V} - 0.000667 \, \text{A} \times 50 \, \Omega \)

\( R = \frac{3 \, \text{V} - 0.000667 \, \text{A} \times 50 \, \Omega}{0.000667 \, \text{A}} \)

Calculating this expression gives:

\( R \approx 4,449.25 \, \Omega \)

Rounding this value, we get: 4,450 Ω.

Answer 2

Let \(G\) be the resistance of the galvanometer, \(V\) be the battery voltage, and \(R\) be the series resistance.

Given:

• \(G = 50 \, \Omega\)
• \(V = 3 \, \text{V}\)
• \(R_1 = 2950 \, \Omega\)
• Full-scale deflection = 30 divisions
• New deflection = 20 divisions

The current for full-scale deflection (\(I_{fs}\)) is given by:

\(I_{fs} = \frac{V}{G + R_1} = \frac{3}{50 + 2950} = \frac{3}{3000} = 0.001 \, \text{A} = 1 \, \text{mA}\)

Since the deflection is proportional to the current, the new current (\(I_{new}\)) for 20 divisions is:

\(I_{new} = I_{fs} \times \frac{20}{30} = 0.001 \times \frac{2}{3} = \frac{2}{3000} \, \text{A}\)

Let \(R_2\) be the new series resistance. Then:

\(I_{new} = \frac{V}{G + R_2}\)

\(\frac{2}{3000} = \frac{3}{50 + R_2}\)

Cross-multiplying:

\(2(50 + R_2) = 3 \times 3000\)

\(100 + 2R_2 = 9000\)

\(2R_2 = 8900\)

\(R_2 = 4450 \, \Omega\)

Therefore, the new resistance in series should be 4450 Ω.