Problem 57 - Master Resource Book in JEE Main Physics

Question

The resistance of the shunt required to allow \(2 \%\) of the main current through the galvanometer of \mathrm{\\{} r e s i s t a n c e ~ \(49 \Omega\) is \(\quad\) [Kerala CET 2008] (a) \(1 \Omega\) (b) \(2 \Omega\) (c) \(0.2 \Omega\) (d) \(0.1 \Omega\) (e) 0010

Step-by-Step Solution

Answer

The resistance of the shunt is \(1 \Omega\). Option (a) is correct.

Step 1: Understanding the Problem

We need to find the resistance of the shunt (denoted by \( R_s \)) such that only \(2\%\) of the current flows through the galvanometer, which has a resistance \( R_g = 49 \Omega \). This means that \(98\%\) of the current should flow through the shunt.

Step 2: Set Up the Current Relationship

Let \( I \) be the total current in the circuit. The current through the galvanometer is \( I_g = 0.02I \) and through the shunt is \( I_s = 0.98I \).

Step 3: Use Parallel Resistance Formula

The galvanometer and the shunt are in parallel. Therefore, the voltage across them is the same. Using the formula for parallel circuits, \( I_g \cdot R_g = I_s \cdot R_s \).

Step 4: Substitute Known Values

Substitute \( I_g = 0.02I \) and \( I_s = 0.98I \) into the parallel resistance equation: \( 0.02I \cdot 49 = 0.98I \cdot R_s \).

Step 5: Simplify and Solve for Rs

Cancel \( I \) from both sides of the equation: \(0.02 \cdot 49 = 0.98 \cdot R_s\). Then solve for \( R_s \):\[ R_s = \frac{0.02 \cdot 49}{0.98} = \frac{0.98}{0.98} \Omega = 1 \Omega \]

Step 6: Conclusion

The calculated resistance of the shunt is \( R_s = 1 \Omega \). Therefore, the correct answer is (a) \(1 \Omega\).

Key Concepts

Parallel Circuits

A parallel circuit is one where multiple paths are available for the current to flow. In these circuits, components are connected alongside each other, rather than in a single line. Each component is connected to the same pair of nodes. This allows each component to have the same voltage across it, which is a key feature of parallel circuits.

Understanding this concept is crucial, as it assumes that the main current splits into parts. Specifically, for a galvanometer and a shunt resistor connected in parallel, we can infer that any current flowing through one also sees the same voltage as the other. This is why we use the property of parallel circuits to calculate the current through different components, ensuring that they all have the same voltage.

Current Division

In circuits with components arranged in parallel, the concept of current division is vital. When current flows into a parallel network, it splits to pass through each component.

The division of current is governed by the relative resistances of the components in each path. Generally, a smaller resistance will allow a larger portion of the current to pass through it, based on Ohm's law. For example, in our specific exercise, only 2% of the current goes through the galvanometer due to its higher resistance compared to the shunt.

• Example: With a galvanometer resistance of 49 Ω, only a minor fraction of the current (2%) passes through, while the majority (98%) is directed through the shunt.
• This phenomenon allows us to calculate how much current goes through each path based purely on their resistances.

Resistance Calculation

Resistance calculation in parallel circuits is distinctive because it takes different paths into account. When calculating for resistance in a parallel configuration, it’s essential to understand that the total resistance is not simply the sum of individual resistances. Instead, it is derived from a specific formula.

• For two resistors in parallel, the formula is: \(\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}\).
• This ensures that the equivalent resistance is less than the smallest individual resistance in the circuit.

In this exercise, knowing the relationship between the resistances (49 Ω for the galvanometer and finding a shunt resistor) helps us calculate the shunt resistance that maintains the desired current distribution. We use it to balance how much of the total current flows through each path.

Ohm's Law

Ohm's Law is a fundamental principle used to understand electrical circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. Its mathematical expression is given by the equation \( V = I \cdot R \). This simple relation connects voltage (V), current (I), and resistance (R).

In the exercise, Ohm's Law helps establish the relationship between the current flowing through the galvanometer and the shunt resistor. Both components have the same voltage across them, being parallel, which is why their voltage equations equate in our calculations.

• For the galvanometer: \(V = 0.02I \cdot 49\)
• For the shunt: \(V = 0.98I \cdot R_s\)

This relation is pivotal when solving the problem, proving indispensable in finding the necessary shunt resistance to achieve the desired current division.