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A galvanometer of resistance 50 Omega i...

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

A

4450 `Omega`

B

5050 `Omega`

C

5550 `Omega`

D

6050 `Omega`

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The correct Answer is:
To solve the problem, we need to determine the new series resistance required to reduce the galvanometer deflection from 30 divisions to 20 divisions. Here’s a step-by-step solution: ### Step 1: Understand the Initial Setup We have a galvanometer with a resistance \( G = 50 \, \Omega \) connected in series with a resistance \( R = 2950 \, \Omega \) and a battery of voltage \( V = 3 \, V \). ### Step 2: Calculate the Current for Full Scale Deflection The current \( I_1 \) that produces full-scale deflection (30 divisions) can be calculated using Ohm's law: \[ I_1 = \frac{V}{R + G} \] Substituting the values: \[ I_1 = \frac{3}{2950 + 50} = \frac{3}{3000} = 0.001 \, A \, (or \, 1 \, mA) \] ### Step 3: Set Up the Equation for Reduced Deflection To achieve a deflection of 20 divisions, we denote the new series resistance as \( R' \). The current \( I_2 \) for this deflection can be expressed as: \[ I_2 = \frac{V}{R' + G} \] Since the deflection is proportional to the current, we can set up the ratio: \[ \frac{I_2}{I_1} = \frac{20}{30} = \frac{2}{3} \] ### Step 4: Relate \( I_2 \) and \( I_1 \) From the above ratio, we can express \( I_2 \): \[ I_2 = \frac{2}{3} I_1 = \frac{2}{3} \times 0.001 = \frac{2}{3000} \, A \] ### Step 5: Substitute \( I_2 \) into the Current Equation Now substituting \( I_2 \) into the equation: \[ \frac{2}{3000} = \frac{3}{R' + 50} \] ### Step 6: Cross-Multiply and Solve for \( R' \) Cross-multiplying gives: \[ 2(R' + 50) = 3 \times 3000 \] \[ 2R' + 100 = 9000 \] \[ 2R' = 9000 - 100 = 8900 \] \[ R' = \frac{8900}{2} = 4450 \, \Omega \] ### Step 7: Conclusion The new series resistance \( R' \) required to reduce the deflection to 20 divisions is \( 4450 \, \Omega \).
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