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

Options

A. `6050 Omega `
B. `4450 Omega `
C. `5050 Omega `
D. `5550 Omega `

Explanation

To solve the problem, we need to determine the series resistance required to reduce the deflection of the galvanometer from 30 divisions to 20 divisions. Here's the step-by-step solution:

### Step 1: Calculate the initial current (I_G) for full-scale deflection
The total resistance in the circuit when the galvanometer shows full-scale deflection (30 divisions) is the sum of the galvanometer resistance (R_G) and the series resistance (R_S):

\[ R_{total} = R_G + R_S = 50 \, \Omega + 2950 \, \Omega = 3000 \, \Omega \]

Using Ohm's Law, the current (I_G) through the galvanometer can be calculated as:

\[ I_G = \frac{V}{R_{total}} = \frac{3 \, V}{3000 \, \Omega} = 1 \times 10^{-3} \, A = 1 \, mA \]

### Step 2: Calculate the current (I') for reduced deflection
To find the current required for 20 divisions, we can set up a proportion based on the full-scale deflection:

\[ I' = \frac{I_G}{30} \times 20 = \frac{1 \times 10^{-3} \, A}{30} \times 20 = \frac{2}{3} \times 10^{-3} \, A = \frac{2}{3} \, mA \]

### Step 3: Set up the equation for the new total resistance
The voltage across the circuit remains the same (3 V). Using Ohm's Law again for the new current (I'):

\[ V = I' \times R_{new} \]

Where \( R_{new} \) is the new total resistance. Thus,

\[ 3 \, V = \left(\frac{2}{3} \times 10^{-3} \, A\right) \times R_{new} \]

Rearranging gives:

\[ R_{new} = \frac{3 \, V}{\left(\frac{2}{3} \times 10^{-3} \, A\right)} = \frac{3 \times 3}{2 \times 10^{-3}} = 4500 \, \Omega \]

### Step 4: Calculate the required series resistance
The new total resistance (4500 Ω) includes the galvanometer resistance (50 Ω). Therefore, the required series resistance (R_S') is:

\[ R_S' = R_{new} - R_G = 4500 \, \Omega - 50 \, \Omega = 4450 \, \Omega \]

### Conclusion
The series resistance that should be connected to reduce the deflection to 20 divisions is **4450 Ω**.

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