Problem 122 - A Complete Resource Book in Physics for JEE Main

Question

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The balance length when key K1 is closed and K2 is open is 12 cm, which matches with the Option (C).

Step 1: Calculate the Potential Gradient when both keys are open

The potential gradient when both the keys are open is given by the EMF(E1) of the battery divided by the length of the potentiometer wire (AB). As such, \( \frac{E1}{AB}=4/50=0.08 V/cm \). Here, the length of wire AB is 50 cm and the EMF E1 of the battery is 4V.

Step 2: Calculate the resistance when both keys are closed

When both keys are closed; R1, R2 and RAB are all connected in parallel. Using the formula for resistance \( R = \frac{1}{\frac{1}{R1} + \frac{1}{R2} + \frac{1}{RAB}}, find the equivalent resistance = R_l which equals \(\frac{1}{\frac{1}{15} + \frac{1}{5} + \frac{1}{10}} \approx 2.5 \Omega\)

Step 3: Calculate the Potential Gradient when both keys are closed

The potential gradient under this scenario can be calculated using the formula \( \frac{E1 \cdot R_l}{R_l+AB}=\frac{4 \cdot 2.5}{2.5 + 10} \approx 0.066667 V/cm \).

Step 4: Calculate the Length when K2 is open and K1 is closed

Finally, when K1 is closed and K2 is open, we find that the equivalent resistance = \( \frac{1}{\frac{1}{R1} + \frac{1}{RAB}} \approx 3.75 \Omega \). Therefore, we have to divide \( \frac{E1 \cdot R_l}{R_l + AB} \) by the equivalent resistance to get the length. Thus, the length becomes \( \frac{4 \cdot 3.75}{3.75 + 10} = 0.12 \) m, or 12 cm.

Key Concepts

Potential Gradient

The potential gradient is a fundamental concept in potentiometer physics problems. It refers to the rate at which the electric potential changes along the length of a conductor or a potentiometer wire.

In the exercise provided, when both keys are open, the potential gradient is determined by dividing the EMF (\(E_1\text{, which is 4V}\)) of the battery by the total length of the potentiometer wire (\(AB\text{, which is 50cm}\)), resulting in \(0.08 V/cm\). The potential gradient essentially tells us how many volts are 'dropped' per centimeter of the wire.

Understanding the potential gradient is crucial for solving potentiometer problems, as it allows one to calculate different voltages at various points along the wire. For students tackling JEE Main Physics questions, mastering this concept can make a significant difference in problem-solving ability.

Ohm's Law

Ohm's Law is a pillar of electronic circuit analysis and a key concept in understanding potentiometer problems. It states that the current (\(I\text{ in Amperes}\)) through a conductor between two points is directly proportional to the voltage (\(V\text{ in Volts}\)) across the two points and inversely proportional to the resistance (\(R\text{ in Ohms}\)) of the conductor. Mathematically, it is expressed as \(I = \frac{V}{R}\).

This principle is crucial when determining the behavior of circuits connected to a potentiometer, such as when calculating the equivalent resistance in various configurations or understanding how voltage dividers work. For JEE Main Physics aspirants, having a strong grasp of Ohm's Law is essential for tackling a wide range of electricity and magnetism problems.

Series and Parallel Resistor Circuits

In the context of the given potentiometer exercise, it is important to understand the concepts of series and parallel resistor circuits to calculate the resistance under different key positions. When resistors are connected end to end, it is a series connection, and their resistances simply add up.

In a parallel connection, as in step 2 of our exercise, resistors are connected across the same two points and the equivalent resistance can be found using the formula \( R = \frac{1}{\frac{1}{R1} + \frac{1}{R2} + \frac{1}{RAB}} \) where \(R1\), \(R2\), and \(RAB\) are the individual resistances. Knowing how to derive and apply the correct formula for series and parallel circuits is critical for resolving complex circuit analysis and is a frequent component of JEE Main Physics examinations.

JEE Main Physics

JEE Main Physics covers a broad range of topics, including the study of potentiometers in the section on Current Electricity. Within this context, understanding how to manipulate and interpret circuits involving potentiometers is vital. Students must be familiar with calculating potential gradients, using Ohm's Law, and solving problems involving series and parallel resistor circuits.

To excel in JEE Main Physics, practice with potentiometer physics problems will enhance both problem-solving speed and accuracy. Such exercises provide valuable experience in circuit analysis, an essential skill for many related questions on the exam. Additionally, grasping the conceptual foundations, like those illustrated in the potentiometer problem provided, is indispensable for success in JEE Main and other competitive exams in physics.