Problem Types & Solved Examples Current Electricity

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Type 1 Pattern — Direct Formula Problems These appear in CBSE boards (1-2 marks), NEET (50% of Current Electricity questions), and JEE Main (warm-up questions). Usually solved in <60 seconds. Speed and zero-error execution is the skill here.

Problem 1 — Drift Velocity

NEET 2022 Easy

Time: ~45 sec

Question:

A copper wire of cross-sectional area 1 mm² carries a current of 1 A. Given that the number density of free electrons in copper is 8×10²⁸/m³, find the drift velocity of electrons.

Given

A = 1 mm² = 10⁻⁶ m²

I = 1 A

n = 8×10²⁸ /m³

e = 1.6×10⁻¹⁹ C

What Examiner Tests

Direct application of I = nAev_d

Unit conversion: mm² → m²

Rearranging for v_d

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Concept Selection: Use I = nAev_d → v_d = I/(nAe). Don't overthink — it's direct substitution.

Solution:

v_d = I/(nAe)

v_d = 1 / (8×10²⁸ × 10⁻⁶ × 1.6×10⁻¹⁹)

v_d = 1 / (8 × 1.6 × 10²⁸⁻⁶⁻¹⁹)

v_d = 1 / (12.8 × 10³) = 1/12800

v_d ≈ 7.8 × 10⁻⁵ m/s

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Shortcut Insight: Remember the order of magnitude: v_d for copper ≈ 10⁻⁴ m/s. If your answer is way off this range, recheck unit conversion of A.

Problem 2 — Resistivity & Resistance

CBSE 2023 Easy

Question:

A wire of resistivity ρ is stretched to double its length. How does its resistance change? (Volume remains constant)

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Concept Selection: R = ρL/A. When length doubles, volume conserved → Area halves. Substitute both changes.

Initial: R₁ = ρL/A

Volume constant: LA = L'A' → (2L)A' = LA → A' = A/2

New resistance: R₂ = ρ(2L)/(A/2) = ρ(2L)(2/A) = 4ρL/A = 4R₁

R₂ = 4R₁ (Resistance becomes 4 times)

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Shortcut: If length becomes n times → R becomes n² times (when volume is constant). Stretched to 3× → R becomes 9×. This is a very common CBSE + JEE Main question pattern.

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Type 2 Pattern — Conceptual Problems These test understanding, not calculation. Common in NEET (assertion-concept), JEE Main MCQ, and CBSE 3-mark questions. Students who only memorize formulas fail here. Think before calculating.

Problem 3 — Temperature Effect (Conceptual)

NEET 2021 Medium

Question:

A metallic conductor and a semiconductor are connected in series. When temperature increases, what happens to the total resistance of the combination?

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Thinking Step: Metallic conductor: R increases with T (positive α). Semiconductor: R decreases with T (negative α). Net effect depends on which change is larger — but for typical materials, semiconductor R decrease dominates at moderate T increase.

In series: R_total = R_metal + R_semi

When T increases:

• R_metal increases (positive α → lattice vibrations increase, τ decreases)

• R_semi decreases (more electron-hole pairs, n increases rapidly)

Net effect: Cannot determine without specific values. The question tests whether you know both effects — answer: "Cannot be determined without knowing magnitudes of α"

If NEET says "total resistance decreases" — they assume semiconductor effect dominates.

Problem 4 — Terminal Voltage (Conceptual)

JEE Main 2019 Medium

Question:

A cell of EMF ε and internal resistance r is connected to external resistance R. The terminal voltage V equals EMF ε. What can you conclude?

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Mistake Alert: Many students say "R is very large." That's partially right. The exact condition is I = 0, which means R → ∞ (open circuit). But the key answer is: "The circuit must be open (no current flowing)."

V = ε − Ir

If V = ε: ε = ε − Ir → Ir = 0 → I = 0 (since r ≠ 0)

∴ Circuit is open (I = 0, R → ∞)

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Exam Trick: Terminal voltage equals EMF ONLY when circuit is open (I=0). During discharge: V < ε. During charging: V > ε.

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Type 3 Pattern — Multi-Step Problems The core of JEE Main and JEE Advanced. These require 3-5 steps and multiple concepts. Missing one step = wrong answer. This is where rank is decided.

Problem 5 — Kirchhoff's Laws (Multi-loop)

JEE Main 2022 Hard

Time: ~4-5 min

Question:

In a circuit, a battery of EMF 12V (internal resistance 1Ω) and another of EMF 8V (internal resistance 2Ω) are connected with resistors R₁=4Ω and R₂=6Ω. The 12V battery drives current clockwise. Using Kirchhoff's laws, find the current through each branch.

Assign branch currents I₁, I₂, I₃

At junction: I₁ = I₂ + I₃ (KCL)

Write KVL for Loop 1 (left loop)

12 − I₁(1) − I₁(4) = 0 (simplified)

Write KVL for Loop 2 (right loop)

Use −8V EMF with its internal resistance

Solve the system simultaneously

Two equations, two unknowns → find currents

Setup: Let I₁ flow from 12V battery (clockwise), I₂ through 8V battery branch, I₃ through common resistor.

KCL at junction A: I₁ = I₂ + I₃

KVL Loop 1: 12 − I₁(1) − I₃(4) = 0 → 12 = I₁ + 4I₃ ...(i)

KVL Loop 2: 8 − I₂(2) − I₃(6) = 0 → 8 = 2I₂ + 6I₃ ...(ii)

From KCL: I₂ = I₁ − I₃

Substituting in (ii): 8 = 2(I₁−I₃) + 6I₃ = 2I₁ + 4I₃ → 4 = I₁ + 2I₃ ...(iii)

From (i)−(iii): 8 = 2I₃ → I₃ = 4 A

From (iii): I₁ = 4 − 2(4) = −4 A (negative → direction was assumed wrong, actual is opposite)

I₂ = I₁ − I₃ = −4 − 4 = −8 A (checks out)

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Key Insight: Negative current means actual direction is opposite to assumed. This is not an error — it's the method working correctly. Never say the answer is wrong if you get negative current.

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Type 4 Pattern — Graph-Based Problems Graph problems appear heavily in NEET (2-3 per year) and JEE Main. They test interpretation of V-I graphs, R-T graphs, and drift velocity graphs. Key skill: identify the slope and its physical meaning.

V-I Graph Analysis

Slope of V-I = R (steeper = higher R)

Linear V-I = ohmic (constant R)

Curved V-I = non-ohmic (R varies)

I-V graph: slope = 1/R (conductance)

R-T Graph Analysis

Metals: positive slope (R increases with T)

Semiconductors: negative slope (R decreases)

Alloys (Nichrome): nearly flat R-T curve

Superconductors: R drops to zero at T_c

Graph Problem — JEE Main Style

Q: The V-I graph for two resistors A and B are straight lines. The slope of A's line is steeper than B's. Which has higher resistance, and which dissipates more power when connected to the same voltage?

Part 1: Slope of V-I = R. Steeper slope = higher R. Therefore, A has higher resistance.

Part 2: At same voltage V: P = V²/R. Higher R → lower P. Therefore, B dissipates more power.

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Trick: Don't confuse "steeper V-I" (higher R) with "steeper I-V" (lower R). Always identify which axis is which before answering slope problems.

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Type 5 — Assertion–Reason (NEET & CBSE Pattern) These are 1-mark questions in NEET (2-3 per topic). The trick: evaluate A and R independently first, then check if R is the correct explanation of A. Students fail by guessing instead of reasoning.

Assertion–Reason Set 1

Assertion (A):

The drift velocity of electrons in a metallic conductor decreases when temperature increases.

Reason (R):

On increasing temperature, the relaxation time τ decreases due to increased lattice vibrations.

A: T, R: T, R explains A

A: T, R: T, R doesn't explain

A: T, R: False

A: False

Answer: Option A (Both True, R is correct explanation of A)

A is TRUE: v_d = eEτ/m. When T increases, τ decreases → v_d decreases (at constant E).

R is TRUE: Higher T → more lattice vibrations → electrons collide more often → τ decreases.

R correctly explains A through the formula v_d = eEτ/m.

Assertion–Reason Set 2

Assertion (A):

Potentiometer is preferred over voltmeter for measuring EMF of a cell.

Reason (R):

A potentiometer draws no current from the cell at balance point.

Answer: Option A (Both True, R correctly explains A)

A is TRUE: Potentiometer measures true EMF, voltmeter reads terminal voltage (slightly less).

R is TRUE: At null deflection, galvanometer shows zero, meaning no current flows from test cell → V_terminal = ε → true EMF is measured.

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Type 6 — Case-Based (CBSE New Pattern from 2021) 5-mark case-based questions in CBSE boards. A passage is given, then 4-5 questions follow. Tests application in context. Many students fail because they don't read the passage carefully.

Case Study — Wheatstone Bridge Experiment

A physics student is performing an experiment using a Meter Bridge to find the unknown resistance of a coil. The standard resistance box is set to R = 10 Ω. The student finds the null point at 40 cm from the left end. She then interchanges the positions of the unknown resistance S and the resistance R, and finds the new null point.

Q1: Find the unknown resistance S. ▼

Using Meter Bridge formula: S = R(100−l)/l

S = 10 × (100−40)/40 = 10 × 60/40 = 10 × 1.5

S = 15 Ω

Q2: If R and S are interchanged, where will the new null point be? ▼

Now S = 15 Ω is in left arm, R = 10 Ω in right arm.

15/10 = l'/(100−l') → 15(100−l') = 10l' → 1500 = 25l' → l' = 60 cm

New null point at 60 cm

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Pattern: Original null at l₁, after interchange: new null = (100 − l₁). Here 100 − 40 = 60. ✓

Q3: Why does the null point shift? ▼

When R and S are interchanged, the balance condition P/Q = R/S changes to P/Q = S/R (reciprocal). Since the wire length ratio l/(100−l) = R/S changes, the null point shifts to maintain the new balance condition.