Q.36 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 36 Solution

Topic: Current Electricity · Official NTA Booklet Code 45

Question

A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:

✓

Verified Against NTA Final Answer Key

Option (2) : 2.0 A

📚 Current Electricity Medium

📋 Given Parameters & Problem Setup

Applied Supply Voltage (V_AB) 50 V

Left Upper Resistor (R_AC) 1 Ω

Left Lower Resistor (R_AD) 3 Ω

Right Upper Resistor (R_CB) 2 Ω

Right Lower Resistor (R_DB) 4 Ω

Branch CD Zero-resistance connecting wire between C and D

⚡

The 30-Second Exam Hall Shortcut: ⚡ Node-Voltage Kirchhoff Shortcut

⏱️ Save ~45 seconds

CD is an equipotential wire (V_C = V_D). R_left = 1 || 3 = 0.75 Ω, R_right = 2 || 4 = 4/3 Ω ⟹ R_eq = 25/12 Ω ⟹ I = 24 A. V_AC = 18 V ⟹ I_1 = 18 A, I_2 = 16 A ⟹ I_CD = 18 - 16 = 2.0 A!

1 Recognize that branch CD is a shorting wire: Node C and Node D are at the same electric potential (V_C = V_D).

2 Left section has 1 Ω and 3 Ω in parallel: R_left = (1 × 3) / (1 + 3) = 3/4 Ω = 0.75 Ω.

3 Right section has 2 Ω and 4 Ω in parallel: R_right = (2 × 4) / (2 + 4) = 8/6 = 4/3 Ω.

4 Total equivalent resistance: R_eq = 3/4 + 4/3 = 25/12 Ω.

5 Total current from 50 V battery: I_total = 50 / (25/12) = 24 A.

6 Voltage across left parallel pair: V_left = 24 A × (3/4 Ω) = 18 V.

7 Current from A to C through 1 Ω resistor: I_AC = 18 V / 1 Ω = 18 A.

8 Voltage across right parallel pair: V_right = 50 V - 18 V = 32 V.

9 Current from C to B through 2 Ω resistor: I_CB = 32 V / 2 Ω = 16 A.

10 By Kirchhoff’s Current Law at node C: Current into CD is I_CD = I_AC - I_CB = 18 A - 16 A = 2.0 A!

11 Select Option (2) with 100% confidence!

🛡️

Why You Can 100% Trust This Shortcut: Kirchhoff’s Current and Voltage Laws (NCERT Physics · Current Electricity: ΣI = 0, ΣV = 0).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Identify Circuit Topology and Equipotential Nodes

Branch CD is an ideal wire with zero resistance connecting junction C and junction D. Therefore, junctions C and D form a single equipotential node:

V_C = V_D = V_{CD}

• The 1 Ω and 3 Ω resistors are connected in parallel between node A and node CD.
• The 2 Ω and 4 Ω resistors are connected in parallel between node CD and node B.

Step 2

Calculate Equivalent Resistances and Total Current

Calculate the equivalent resistance of each section and the whole network:

R_{left} = \frac{1 \times 3}{1 + 3} = \frac{3}{4}\,\Omega, \quad R_{right} = \frac{2 \times 4}{2 + 4} = \frac{4}{3}\,\Omega \implies R_{eq} = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}\,\Omega

• Total current delivered by the 50 V battery:
• I_{total} = \frac{50}{25/12} = 50 \times \frac{12}{25} = 24\text{ A}.

Step 3

Calculate Currents and Apply KCL at Node C

Determine the potential at the central node and find currents entering and leaving node C:

V_A - V_C = I_{total} R_{left} = 24 \times \frac{3}{4} = 18\text{ V} \implies I_{AC} = \frac{18\text{ V}}{1\,\Omega} = 18\text{ A}

• V_C - V_B = 50 - 18 = 32 V ⟹ I_{CB} = 32 V / 2 Ω = 16 A.
• KCL at node C: I_{AC} = I_{CB} + I_{CD} ⟹ I_{CD} = 18 A - 16 A = 2.0 A (flowing downward from C to D).
• Verification at node D: I_{AD} = 18 V / 3 Ω = 6 A; I_{DB} = 32 V / 4 Ω = 8 A; 6 A + 2 A = 8 A (verified!).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Calculation error in node voltage.

⚠️

Student Fallacy (Why candidates pick this trap): Approximation error in current division.

2 B

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. Rigorous nodal analysis yields exactly 2.0 A.

💡

Why this is airtight: None; perfect execution of KCL.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arithmetic slip.

⚠️

Student Fallacy (Why candidates pick this trap): Using wrong equivalent resistance.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Obtained if total current is calculated incorrectly as 30 A.

⚠️

Student Fallacy (Why candidates pick this trap): Rounding total resistance to 2 Ω.

⚠️

NTA Trap Alert: Treating Network as an Unbalanced Wheatstone Bridge

Some candidates treat CD as a galvanometer branch in an unbalanced bridge and write lengthy 3-loop mesh equations. Because CD is a plain ideal wire, V_C = V_D immediately simplifies the network into two parallel pairs in series!

🛡️ How to avoid the -1 mark: Whenever a branch has zero resistance, combine the connected nodes into one single node!

📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Current Electricity Standard Benchmark

Core Scientific Principle Tested:

"Σ I_{in} = Σ I_{out} at any junction; shorted branches create equipotential nodes."

Syllabus Focus: Kirchhoff’s Current Law & Nodal Analysis

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

V_CD = 32 V ⟹ I_in = 18 A, I_out = 16 A ⟹ Difference = 2.0 A through CD!

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NEET 2025 Physics Question 36 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Node-Voltage Kirchhoff Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Identify Circuit Topology and Equipotential Nodes

Calculate Equivalent Resistances and Total Current

Calculate Currents and Apply KCL at Node C

360° Option Elimination Engine Key Feature

NTA Trap Alert: Treating Network as an Unbalanced Wheatstone Bridge

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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