Скачать или смотреть JEE Mains 2026 (24 Jan Shift 1) Physics Solution: Current Electricity & Voltmeter Resistance

Welcome to Physics Unlocked by Kanak Yadav! In this video, we solve a Current Electricity problem from the JEE Mains 2026 exam (24 January, Shift 1).

📌 Question:
Two resistors of 100 Ω each are connected in series with a 9 V battery. A voltmeter of 400 Ω resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ________ V.

Options:
1. 4
2. 3
3. 4.5
4. 2

⏱️ Timestamps:
0:00 - Problem Statement
0:25 - Circuit Diagram & Setup
1:03 - Calculating Effective Resistance
1:55 - Main Branch Current Calculation
2:13 - Current Division (KCL & Ohm's Law)
2:58 - Final Voltmeter Reading Calculation
3:11 - Outro

📝 Detailed Step-by-Step Solution:
Let the two resistors be R1 = 100 Ω and R2 = 100 Ω.
The battery voltage is V = 9 V.
The voltmeter has a resistance Rv = 400 Ω and is connected in parallel with R1.

Step 1: Find the effective resistance of the parallel combination (R1 and Rv).
1 / Rp = (1 / R1) + (1 / Rv)
1 / Rp = (1 / 100) + (1 / 400)
1 / Rp = (4 / 400) + (1 / 400) = 5 / 400
Rp = 400 / 5 = 80 Ω

Step 2: Find the total equivalent resistance (Req) of the circuit.
The parallel combination (Rp) is in series with R2.
Req = Rp + R2
Req = 80 Ω + 100 Ω = 180 Ω

Step 3: Calculate the main branch current (I) from the battery.
I = V / Req
I = 9 V / 180 Ω = 0.05 A

Step 4: Find the current flowing through the voltmeter (i1).
The main current (0.05 A) divides into i1 (through the 400 Ω voltmeter) and i2 (through the 100 Ω resistor).
Using Kirchhoff's Current Law (KCL): i1 + i2 = 0.05 A
Since they are in parallel, the voltage across both branches is equal:
400 * i1 = 100 * i2
i2 = 4 * i1

Substitute i2 back into the KCL equation:
i1 + 4 * i1 = 0.05
5 * i1 = 0.05
i1 = 0.01 A

Step 5: Calculate the voltmeter reading.
The reading is the voltage drop across the 400 Ω resistance.
V_reading = i1 * Rv
V_reading = 0.01 A * 400 Ω = 4 V

The correct option is 1.

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