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PhysicsVery High Weightage ★★★★★Class 12

Current Electricity

Ohm's law, Kirchhoff's laws, Wheatstone bridge, potentiometer, and cells — the most numerical chapter in electricity. Expect 4–5 EAPCET questions.

4–5Questions in EAPCET
~5%Paper Weightage
12Core Formulas
4Mistake Traps

Concept Core

From Ohm's Law to Kirchhoff's laws — mastering circuits.

Ohm's Law & Resistance

V = IR. Resistance R = ρL/A (ρ = resistivity, L = length, A = cross-section).

Series: R_eff = R₁ + R₂ + ... (same current; voltage divides) Parallel: 1/R_eff = 1/R₁ + 1/R₂ + ... (same voltage; current divides)

Temperature: R = R₀(1 + αT) where α = temperature coefficient of resistance. For metals α > 0 (resistance increases with T). For semiconductors/NTC: α < 0.

Kirchhoff's Laws

KCL (Junction Rule): Sum of currents entering a junction = sum of currents leaving. ΣI_in = ΣI_out (conservation of charge)

KVL (Loop Rule): Sum of EMFs and potential drops around any closed loop = 0. ΣE = ΣIR (conservation of energy)

Wheatstone Bridge

For a balanced Wheatstone bridge: no current through galvanometer, and:

P/Q = R/S   (balanced condition)

Applications: measurement of unknown resistance. Metre bridge uses a uniform wire: P/Q = l/(100−l) where l is balance point length from left end.

EMF, Internal Resistance & Cells

A cell with EMF ε and internal resistance r:

Terminal voltage V = ε − Ir (discharging) V = ε + Ir (charging) Current I = ε/(R+r)

Cells in series: ε_eff = ε₁+ε₂, r_eff = r₁+r₂

Cells in parallel: ε_eff = ε₁ = ε₂ (if equal), r_eff = r/n

Power & Electrical Energy
P = VI = I²R = V²/R Energy = Pt = VIt = I²Rt

Maximum power transfer: Maximum power delivered to external resistance R when R = r (internal resistance). P_max = ε²/(4r)

Potentiometer

Used to compare EMFs and find internal resistance. Key principle: no current is drawn when balance is achieved → accurate measurement.

ε₁/ε₂ = l₁/l₂ (balance lengths) r = R(l₁−l₂)/l₂ (internal resistance)

Formula Vault

All current electricity formulas — circuits, cells, and measurements.

Ohm's Law
V = IR
V in volts; I in amps; R in ohms
Resistance
R = ρL/A
ρ = resistivity; L = length; A = area
Series Resistors
R_eff = R₁ + R₂ + ...
Same current; voltage divides
Parallel Resistors
1/R_eff = 1/R₁ + 1/R₂
Same voltage; current divides
Cell Terminal Voltage
V = ε − Ir (discharging)
V = ε + Ir when charging
Cell Current
I = ε/(R + r)
R = external resistance
Wheatstone Balance
P/Q = R/S
No current through galvanometer
Metre Bridge
P/Q = l/(100−l)
l = balance length from left
Power
P = VI = I²R = V²/R
Three equivalent forms
Max Power Transfer
P_max = ε²/4r when R=r
R = r for maximum power to load
Potentiometer EMF ratio
ε₁/ε₂ = l₁/l₂
l = balance length
KVL
ΣE = ΣIR around loop
Signs: EMF and IR in same direction → positive

Worked Examples

5 problems — series/parallel, cell with internal resistance, Wheatstone, KVL, and power.

EasyThree resistors 2Ω, 3Ω, 6Ω in parallel — find equivalent
Find equivalent resistance of 2Ω, 3Ω, and 6Ω resistors in parallel.
1
1/R_eff = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
2
R_eff = 1 Ω
✓  R_eff = 1 Ω
EasyCell with ε=12V, r=2Ω, R=10Ω — find terminal voltage
A cell of EMF 12V and internal resistance 2Ω is connected to 10Ω external. Find terminal voltage.
1
I = ε/(R+r) = 12/12 = 1 A
2
Terminal voltage V = ε − Ir = 12 − 1×2 = 10 V
✓  Terminal voltage = 10 V
MediumWheatstone bridge: P=10Ω, Q=20Ω, R=15Ω — find balanced S
In a Wheatstone bridge, P = 10Ω, Q = 20Ω, R = 15Ω. Find S for balance.
1
Balance condition: P/Q = R/S → S = QR/P = 20×15/10 = 30 Ω
✓  S = 30 Ω
EAPCET LevelApply KVL to find current in a loop with two cells
In a loop: cell 1 (ε=10V, r=1Ω) and cell 2 (ε=4V, r=1Ω) connected in series with R=5Ω. Cells oppose each other. Find current.
1
Assign current I clockwise. Apply KVL around loop:
2
10 − 4 − I(1) − I(5) − I(1) = 0
3
6 − 7I = 0 → I = 6/7 ≈ 0.86 A
✓  I = 6/7 A ≈ 0.86 A
Trap QuestionMaximum power to external resistance when R = 0 (short circuit)?
A student claims maximum power to external R occurs when R = 0 (short circuit current maximises I). Is this correct?
1
The trap: Maximum current (at R=0) doesn't mean maximum power to R.
2
P_external = I²R = [ε/(R+r)]² × R
3
At R=0: P = 0 (zero voltage across zero resistance)
4
At R=∞: P = 0 (zero current through infinite resistance)
5
Maximum P occurs when dP/dR = 0 → solving gives R = r.
6
P_max = ε²/(4r) when R = r (external matches internal resistance)
✓  Maximum power when R = r (not R = 0); P_max = ε²/(4r)

Mistake DNA

4 circuit errors from EAPCET distractor analysis.

🔋
Applying Series Formula to Parallel Resistors (And Vice Versa)
The most common circuit error. Series and parallel combination formulas are frequently swapped.
❌ Wrong
Parallel: R = R₁+R₂ ✗ (that's series!) Series: 1/R=1/R₁+1/R₂ ✗
✓ Correct
Series: R = R₁+R₂ ✓ Parallel: 1/R=1/R₁+1/R₂ ✓ Check: parallel R < smallest R
Quick check: parallel resistance is ALWAYS less than the smallest individual resistance. If your answer is larger, you used the wrong formula.
Terminal Voltage = EMF (Ignoring Internal Resistance)
A real cell has internal resistance r. Terminal voltage is less than EMF when current flows.
❌ Wrong
Terminal voltage =
EMF = 12V ✗ (only when I=0 or r=0)
✓ Correct
V = ε − Ir ✓ = 12 − I×r < 12 V = ε only when open circuit (I=0)
Terminal voltage equals EMF only when the circuit is open (no current). When current flows, voltage drops across internal resistance r.
🔌
KVL: Wrong Signs for EMF and Resistance
In KVL, sign conventions for EMF and IR drops must be consistent with the chosen loop direction.
❌ Wrong
Traversing against current
I: add +IR (wrong sign) ✗
Traversing into − terminal of cell: add +ε ✗
✓ Correct
Against current: −IR ✓ With current: +IR ✓ Into + terminal: +ε ✓ Into − terminal: −ε ✓
KVL sign convention: moving in direction of assumed current → voltage drop −IR. Moving from − to + inside the cell → EMF +ε. Be consistent with loop direction.
⚖️
Max Power Transfer: Maximum Current = Maximum Power to R
P_external = I²R. Maximum I (when R=0) gives I_max but P = I²×0 = 0. Maximum P requires balancing I and R.
❌ Wrong
Max P at R=0 (short circuit) I = ε/r maximum ✗ P = I²×0 = 0 W ✗
✓ Correct
Max P_external when R=r ✓ P_max = ε²/(4r) ✓ Derivative condition: dP/dR = 0 → R = r
Power to R is P = [ε/(R+r)]²×R. This is zero at R=0 and R=∞, maximum somewhere in between. Differentiating and setting to zero gives R = r.

Chapter Intelligence

Current electricity is the most tested chapter in Class 12 Physics — numerical mastery is essential.

EAPCET Weightage (2019–2024)
Resistor combinations
~9
Cell with internal resistance
~7
KVL circuit analysis
~6
Wheatstone bridge / metre bridge
~5
Power in circuits
~4
Potentiometer
~3
High-Yield PYQ Patterns
Equivalent resistance of combinationTerminal voltage of cellFind R for balanced bridgeKVL: current in two-cell loopMaximum power condition R=rPotentiometer balance length ratioResistance from resistivity formula
Exam Strategy
  • Circuit simplification: first identify series/parallel relationships visually. Redraw the circuit if needed. Work from innermost combination outward.
  • Cell problems: always use V = ε − Ir (not V = ε). The voltage drop Ir across internal resistance is often the key to the answer.
  • KVL: choose a loop direction, apply the sign convention consistently. Sum = 0. Two unknowns → two loops → two equations.
  • Maximum power transfer: R = r. This appears as both a formula question and a conceptual question. Know both forms: P_max = ε²/(4r).
  • Current electricity links to Magnetism (magnetic force on current-carrying conductor) and Electromagnetic Induction (changing current creates changing B → EMF).