PYQ Analysis — Last 10 Years
Know what's been asked, how often, and what patterns repeat. Stop guessing. This is data-driven preparation.
47
JEE Main Qs (10yr)
18
JEE Adv Qs (10yr)
22
NEET Qs (10yr)
35
CBSE PYQs
JEE Main
JEE Main — 10 Year Analysis (2014–2024)
🔬 Key Finding
JEE Main asks 3–5 capacitor questions per year across both sessions. Energy stored and combination circuits are the most repeated topics. Since 2020, at least 1 RC circuit question appears per year.
Topic-wise Weightage (JEE Main)
Combination of Capacitors
88%
Energy Stored
82%
Parallel Plate Capacitor
75%
Dielectric Effects
70%
RC Circuits
55%
Charge Redistribution
50%
Spherical/Cylindrical Cap
30%
Year-wise Question Map
2024
4 questions
2023
5 questions
2022
3 questions
2021
4 questions
2020
5 questions
2019
4 questions
2018
3 questions
2017
4 questions
Solved PYQ Examples
A capacitor of 500 μF is charged to 100 V. The energy stored is then used to charge another 500 μF capacitor. What is the energy stored in the second capacitor?
1
Initial energy in C₁: U₁ = ½×500×10⁻⁶×100² = 2.5 J
2
When C₁ (charged) is connected to C₂ (uncharged), charge redistributes: V_final = (C₁V₁)/(C₁+C₂) = 500×100/(500+500) = 50 V
3
Energy in C₂: U₂ = ½×500×10⁻⁶×50² = 0.625 J
⚡ Note: U₂ = U₁/4. When charge is shared equally between two identical capacitors, energy is HALVED (factor 1/2 from each). Energy loss = 1.25 J (heat)
In an RC circuit, R=2MΩ, C=5μF. The capacitor is charged to 10V. Find the time constant and the time after which charge reduces to 1/e of initial.
1
τ = RC = 2×10⁶ × 5×10⁻⁶ = 10 s
2
q = Q₀e^(-t/τ). For q = Q₀/e: e^(-t/τ) = 1/e → t/τ = 1 → t = τ = 10 s
⚡ By definition, the time constant τ is the time for charge to fall to 1/e (≈37%) of initial value during discharge.
JEE Advanced
JEE Advanced — 10 Year Analysis (2014–2024)
🔬 Advanced Pattern Insight
JEE Advanced capacitor problems are always multi-concept. They're rarely standalone — expect capacitors combined with: RC + switch problems, conductor mechanics, dielectric force problems, or Wheatstone-type networks. Integer type has appeared 8/10 years.
Topic-wise Complexity (JEE Adv)
Circuit Networks (Kirchhoff)
80%
Switch + RC + Transient
70%
Energy Accounting (Battery)
65%
Force / Work on plates/dielectric
55%
Charge distribution in networks
50%
In a circuit: C₁=C₂=C₃=C, all connected in a triangular loop with a battery V across one branch. Find charge on each capacitor.
🧠 Approach: Redraw + KVL + Charge Conservation
Label charges q₁, q₂, q₃ on each capacitor. Use KVL for two independent loops + charge conservation at the floating node.
1
Let charges be q₁, q₂, q₃. At floating node: q₁ = q₂ + q₃ (charge conservation)
2
Loop 1 (battery + C₁ + C₂): V = q₁/C + q₂/C
3
Loop 2 (C₂ + C₃): q₂/C = q₃/C → q₂ = q₃
4
Solve: q₁ = 2q₂, q₁ = (2/3)CV, q₂ = q₃ = CV/3
⚡ Symmetric circuits: if C₂=C₃ and they're in symmetric positions, q₂=q₃ by symmetry. This alone halves the equations needed.
NEET
NEET — 10 Year Analysis (2014–2024)
🎯 NEET Strategy
NEET asks 1–2 capacitor questions. They're almost ALWAYS conceptual (not computational). Focus on: dielectric insertion with/without battery, energy comparison, voltage and charge changes. NEVER a Kirchhoff's law problem in NEET.
Most Repeated NEET Topics
Dielectric Insertion Effects
90%
Energy Stored (formula)
80%
Series/Parallel Combo
65%
Capacitance formula C=ε₀A/d
55%
A parallel plate capacitor is connected to a battery. The battery maintains a constant voltage V = 50 V. A dielectric slab of K = 3 is inserted. Which one is correct?
(A) Charge increases to 3Q₀ (B) Voltage drops to V₀/3 (C) Energy decreases (D) E field decreases to E₀/3
(A) Charge increases to 3Q₀ (B) Voltage drops to V₀/3 (C) Energy decreases (D) E field decreases to E₀/3
🧠 Battery connected → V = constant
C → 3C₀, Q = CV → Q → 3Q₀ (increases), V = constant = 50V, E = V/d = constant, U = ½CV² → 3U₀ (increases)
✓
Answer: (A) — Charge increases to 3Q₀. All others are wrong.
❌ NEET 2022 Trap Analysis
62% of students chose (D) — they confused battery-connected with battery-disconnected. With battery connected, E = V/d = constant. E only decreases when Q is fixed (battery disconnected).
Two capacitors of 2μF and 3μF are connected in parallel, then this combination is connected in series with a 5μF capacitor. Find the equivalent capacitance.
1
Parallel: C_p = 2+3 = 5 μF
2
Series with 5μF: C = (5×5)/(5+5) = 25/10 = 2.5 μF
⚡ Two equal capacitors in series = half the value. 5μF in series with 5μF = 2.5μF. Simple pattern to remember.
CBSE Board
CBSE Board — Analysis (2015–2024)
🔬 CBSE Exam Pattern
Capacitors in CBSE boards: 2 marks (direct formula/definition), 3 marks (derivation or 2-3 step numerical), 5 marks (full derivation + numerical). The parallel plate capacitor derivation has appeared in 7 out of last 10 years.
| Year | 1-2 Mark Q | 3 Mark Q | 5 Mark Q | Total Marks |
|---|---|---|---|---|
| 2024 | Capacitance definition | Series combination derivation | Parallel plate + dielectric derivation | 10 |
| 2023 | Energy formula | Redistribution of charge | Spherical capacitor derivation | 10 |
| 2022 | Dielectric constant | Energy stored numerical | Parallel plate derivation | 10 |
| 2021 | Capacitors in parallel | Circuit numerical | - | 5 |
| 2020 | Definition | Combination circuit | Derivation + numerical | 10 |
| 2019 | SI unit of capacitance | Series derivation | Parallel plate + energy | 10 |
CBSE 5-Mark: Parallel Plate Capacitor with Dielectric Slab
How to write perfect 5-mark answer
1
Circuit diagram: Draw capacitor with dielectric slab of thickness t, air gap (d-t)
2
Setup: Treat as two capacitors in series — C_air = ε₀A/(d-t) and C_dielectric = Kε₀A/t
3
Series combination: 1/C = 1/C_air + 1/C_dielectric = (d-t)/(ε₀A) + t/(Kε₀A)
4
Final formula: C = ε₀A / (d - t + t/K)
5
Special case: If t = d (full slab): C = Kε₀A/d = KC₀ ✓
🎯 CBSE Marking Tip
Draw the diagram (1 mark), setup equations (1 mark), combine correctly (1 mark), final formula (1 mark), verify special case (1 mark). This structure guarantees full marks if done correctly.
Pattern Recognition
Repeated Question Patterns
These patterns repeat every 2-3 years. Recognize them immediately in the exam.
NEET 2016, 2019, 2022
JEE Main 2017, 2020, 2023
This is the most recycled problem type. Formula: V_common = (C₁V₁+C₂V₂)/(C₁+C₂), ΔU = ½(C₁C₂/(C₁+C₂))(V₁-V₂)²
🎯 30-second solver
Identify polarity (same/opposite) → Write V_common formula → Compute Uᵢ and Uf → ΔU = Uᵢ - Uf. Done in 60 seconds max.
NEET every year
JEE Main 2015, 2018, 2022
Always specify: battery connected (V const) or disconnected (Q const). Then fill the comparison table.
JEE Main 3-4 times/year
The trick: always redraw the circuit. Identify which nodes are connected to + and - of battery. All capacitors connected between same two nodes are in PARALLEL.
🧠 Node Identification Trick
Label each node A, B, C... Draw wires as dots. Capacitors connected A-B and A-B are parallel. Capacitors connected A-B and B-C are series (if B is floating). This simplification works for 90% of JEE circuit problems.
JEE Main every year since 2019
Template: q = Q₀(1-e^(-t/τ)) for charging. Set equal to target value, take ln of both sides.
t = τ·ln(Q₀/(Q₀-q))
General charging time formula