🔹 Question:

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance R_p = 1Ω as shown in the figure. An external resistance R_e = 2Ω is connected via the sliding contact. Battery voltage = 0.9 V.Find the current through the circuit.Options:
(1) 0.3 A    (2) 1.35 A    (3) 1.0 A    (4) 0.9 A

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In this article, we will explore the jee mains 2025 potentiometer question in detail.

Understanding the jee mains 2025 potentiometer question is crucial for JEE aspirants.

https://youtube.com/watch?v=PHjC9GACXUg%3Fsi%3DYO6mzgc1K1mEHh8z

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🧠 Step-by-Step Solution:

🔧 Step 1: Understand the Circuit

The jee mains 2025 potentiometer question tests your knowledge of circuits and resistance.

The potentiometer of resistance 1Ω is divided equally because the sliding contact is in the middle.
So we get two 0.5Ω resistors on either side of the sliding contact.
The 2Ω resistor is connected between the sliding contact and one terminal of the potentiometer.

🔧 Step 2: Simplify the Equivalent Circuit

We can redraw the circuit as:

• Top branch: 0.5Ω
• Bottom branch: 0.5Ω + 2Ω = 2.5Ω

These branches are in parallel, forming a loop with the battery.

🔧 Step 3: Calculate the Equivalent Resistance

In tackling the jee mains 2025 potentiometer question, it is essential to apply Ohm’s Law effectively.

Use parallel resistance formula:

Req = 0.5 + (0.5 × 2) / (0.5 + 2)
       = 0.5 + 1 / 2.5
       = 0.5 + 0.4 = 0.9 Ω

🔧 Step 4: Apply Ohm’s Law

This is a common approach for solving the jee mains 2025 potentiometer question.

Now calculate total current using:

I = V / R = 0.9 / 0.9 = 1.0 A

Thus, for the jee mains 2025 potentiometer question, the calculated current is paramount.

The result of your calculations for the jee mains 2025 potentiometer question will guide your understanding.

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✅ Final Answer:

Current in the circuit = 1.0 A    ✅ (Option 3)

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