Problem 1925 - JEE Physics

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) \(5550 \Omega\)

Step-by-Step Solution

Answer

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

Step 1: 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. \(I_{total} = \frac{V}{R_{total}}\) \(I_{total} = \frac{3}{2965}\)

Step 2: 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. \(I_2 = I_{total} \cdot \frac{2}{3}\)

Step 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) \(4450 \Omega\)

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics used to relate the voltage, current, and resistance in a circuit. It is usually expressed with the formula:\[ I = \frac{V}{R} \]Where:

• \(I\) is the current in amperes (A),
• \(V\) is the voltage in volts (V),
• \(R\) is the resistance in ohms (Ω).

Ohm's Law is incredibly useful for calculating how changes in voltage or resistance affect the current in a circuit. In this context, the Galvanometer circuit voltage was known, and we calculated the total resistance to find the current. This application provided insights into how the circuit's behavior would change if we altered any component.

Resistance Calculation

Resistance calculation is crucial in understanding how different components in a circuit interact with each other. In the exercise, we had to find the total resistance of the circuit with the initial setup. The resistance values, given as Galvanometer resistance and a series resistance, combine to determine how much the circuit resists the flow of current.To calculate total resistance in a series circuit, you sum up all individual resistances:\[ R_{total} = R_{galv} + R_{series} \]This calculation is important because it allowed us to determine how the current split between the different parts of the circuit. With an initial resistance of \(2965 \, \Omega\), the resistance needed to change to appropriately adapt to a reduced current, aiming for a deflection of 20 instead of 30 divisions. This requires recalculating and adjusting the resistor in series to achieve the desired current.

Current Divider Rule

The current divider rule helps distribute the current in various branches of a parallel circuit. While primarily used in parallel circuits, its concept helps understand current distribution when altering circuit components in series to achieve specific outcomes.In this exercise, while dealing with a series circuit, we still approached the problem by adjusting resistance values to influence the current's efficacy. The key was understanding the proportion of currents resulting in deflection. Given the original and desired deflections (30 and 20 divisions, respectively), we found the ratio of the two currents:\[ \frac{I_2}{I_1} = \frac{20}{30} = \frac{2}{3} \]Using this ratio, we calculated a new current corresponding to a 20 divisions deflection by multiplying the initial current by \(\frac{2}{3}\). Finally, by adjusting the total resistance to generate this new current, we achieved the desired deflection outcome. Hence, understanding current distribution based on resistance adjustments plays a pivotal role in precision tuning of instruments like a galvanometer.