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.