Question

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a 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 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)

Short answer

The short answer is: (b) \(4450 \Omega\).

Step-by-step solution

Calculate the total current in the circuit with 30 divisions deflection

Using Ohm's Law, we can calculate the total current in the circuit. The formula for Ohm's Law is: \(I = \frac{V}{R}\) Where: I - current (A) V - voltage (V) R - resistance (Ω) We are given the Galvanometer resistance (15Ω) and the additional resistance in series (2950Ω), so we can calculate the total resistance of the circuit. \(R_{total} = R_{galv} + R_{series}\) \(R_{total} = 15 + 2950\) \(R_{total} = 2965\, \Omega\) Now, we can use the voltage of the battery (3V) and the total resistance to calculate the current in the circuit. \(I_{total} = \frac{V}{R_{total}}\) \(I_{total} = \frac{3}{2965}\)

Use current divider formula for 20 divisions deflection

To reduce the deflection from 30 divisions to 20 divisions, we need to find the current ratio that corresponds to this reduction. The current ratio is: \(\frac{I_2}{I_1} = \frac{20}{30}\) \(I_2 = I_1 \cdot \frac{2}{3}\) Now, we can multiply the total current from step 1 by the ratio to find the current \(I_2\) we desire to have in our circuit for 20 divisions deflection. \(I_2 = I_{total} \cdot \frac{2}{3}\)

Find the new resistance in series

Using Ohm's Law, we can calculate the new total resistance required for the desired current (I_2). \(R_{new-total} = \frac{V}{I_2}\) Now, we just need to subtract the Galvanometer resistance from the new total resistance to find the new resistance in series with the Galvanometer to achieve 20 divisions deflection. \(R_{new-series} = R_{new-total} - R_{galv}\) Perform the calculations to find the new resistance in series, rounding to the nearest integer: \(R_{new-series} = \approx 4449 \, \Omega\) So, the correct answer is: (b) \(4450 \Omega\)