Q.28 Physics ⭐ 360° Option Elimination Active
NEET 2023 Physics Question 28 Solution
Topic: Current Electricity · Official NTA Booklet Code G5
Question
The magnitude and direction of the current in the following circuit is
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Verified Against NTA Final Answer Key
Option (3) : 0.5 A from A to B through E
📚 Current Electricity Easy
📋 Given Parameters & Problem Setup
Battery 1 EMF 10 V (connected with 2 Ω internal/series resistance)
Battery 2 EMF 5 V (connected opposing with 1 Ω resistance)
Load Resistor 7 Ω in the lower branch
Total loop resistance (R_tot) 2 Ω + 1 Ω + 7 Ω = 10 Ω
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⏱️ Save ~40 seconds The 30-Second Exam Hall Shortcut: ⚡ Opposing Cells Net EMF Rule: I = (E₁ - E₂) / ΣR
Net EMF = 10 V - 5 V = 5 V (dominated by 10 V cell). Total resistance = 2 + 1 + 7 = 10 Ω. Current I = 5 / 10 = 0.5 A from A to B through E in 15 seconds!
1 Inspect cell orientations: The 10 V and 5 V batteries oppose each other in the single closed loop.
2 The larger battery (10 V) dictates the net EMF: E_net = 10 V - 5 V = 5 V.
3 Sum all resistors in series around the single loop: R_tot = 2 + 1 + 7 = 10 Ω.
4 Calculate current magnitude: I = E_net / R_tot = 5 V / 10 Ω = 0.5 A.
5 Determine direction: The 10 V battery pushes current out of its positive terminal, which directs current from A to B through point E.
6 Select Option (3) 0.5 A from A to B through E!
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Why You Can 100% Trust This Shortcut: Current Electricity · Kirchhoff’s Voltage Law (Net current in a single closed loop equals algebraic sum of EMFs divided by total series resistance).
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Step-by-Step Rigorous Workflow & Mathematical Proof
Step 1
Circuit Identification and Loop Path
The given circuit is a single closed loop ABCDA. All components (the 2 Ω resistor, the 10 V cell, junction E, the 5 V cell, the 1 Ω resistor, and the 7 Ω resistor) are connected in series.
Step 2
Net Electromotive Force (EMF) in the Loop
The positive terminal of the 10 V source faces towards E (pushing current from left to right, i.e., from A towards B), while the 5 V source faces into E with its positive terminal pushing opposite. Because they are in opposition, the net driving EMF is the difference:
Step 3
Calculation of Current Magnitude
By Ohm’s Law / Kirchhoff’s Loop Rule applied to the single closed path:
Step 4
Determination of Current Direction
Since E₁ (10 V) is greater than E₂ (5 V), the net current flows in the direction driven by the 10 V battery: flowing clockwise through the loop, which corresponds to travelling from A to B through point E.
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4-Option Deep Scan
360° Option Elimination Engine Key Feature
Diagnose every choice: Know exactly why the right option works and why distractors fail.
1 (1)
✗ Incorrect Verdict & Scientific Reason:
Incorrect magnitude and wrong direction (result of adding EMFs 10 + 5 = 15 V).
⚠️
Student Fallacy (Why candidates pick this trap): Candidate erroneously treated the opposing cells as aiding cells (15 V / 10 Ω = 1.5 A).
2 (2)
✗ Incorrect Verdict & Scientific Reason:
Wrong magnitude and opposite direction.
⚠️
Student Fallacy (Why candidates pick this trap): Candidate inverted the formula or made an arithmetic error.
3 (3)
✓ 100% Correct Verdict & Scientific Reason:
Accurate magnitude (5 V / 10 Ω = 0.5 A) and correct clockwise direction driven by 10 V battery.
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Why this is airtight: None. Candidate correctly identified opposing polarities.
4 (4)
✗ Incorrect Verdict & Scientific Reason:
Incorrect resistance summation (omitted the 1 Ω internal resistor: 5 / (2 + 7) = 5/9 A).
⚠️
Student Fallacy (Why candidates pick this trap): Candidate forgot the 1 Ω resistor in the loop.
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NTA Trap Alert: Adding EMFs Instead of Subtracting Opposing Batteries
When two cells face each other positive-to-positive or negative-to-negative, their EMFs oppose each other (Net EMF = 10 - 5 = 5 V). Adding them (10 + 5 = 15 V) yields 1.5 A (Option 1).
🛡️ How to avoid the -1 mark: Trace clockwise: if you exit positive of one cell (+10 V) and exit positive of another (-5 V), they oppose!
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Core NEET Syllabus Standard
National Eligibility cum Entrance Test (NEET UG) · Current Electricity Standard Benchmark
Core Scientific Principle Tested:
"In a single series loop, I = ΣE / ΣR, where opposing cells subtract their potentials."
Syllabus Focus: Kirchhoff’s Laws and Grouping of Cells
Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗
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