📖 Chapter PN07 · NDA Class 11–12 Level🎯 NDA Level : High Priority
Electricity is the most extensively tested chapter in NDA Physics — covering the entire spectrum from static charges to current circuits, from Joule heating to electromagnetic induction. It underpins all modern defence technology: from radio communication to radar, from battery-powered field equipment to ship propulsion systems. Average students who master Ohm's law, circuit rules, and the magnetic effect of current can answer a large portion of NDA Physics questions from this single chapter.
📌 What to expect in NDA (based on 2022–2025 pattern): (1) Coulomb's law, electric charge, conductors vs insulators; (2) Ohm's law — V = IR; resistivity; factors affecting resistance; (3) Series and parallel combinations of resistors — equivalent resistance; (4) Joule's law of heating — H = I²Rt; electrical power P = VI; (5) Magnetic field due to current — straight wire (right-hand rule), loop, solenoid; (6) Motor principle — force on current-carrying conductor in magnetic field (F = BIL); (7) Electromagnetic induction — Faraday's law, Lenz's law, AC/DC generators.
Topics at a Glance
① Electrostatics
Charge, Coulomb's law, conductors, insulators, field
② Current Electricity
I, V, R, Ohm's law, resistivity, EMF
③ Circuits & Power
Series/parallel, Joule heating, power, fuse
④ Magnetic Effects & EMI
Current → field, motor, Faraday, Lenz's law
1. Electrostatics
1.1
Electric Charge, Coulomb's Law & Electric Field
The foundation of all electricity — the force between charges
All matter contains electric charge. Positive charge arises from proton excess; negative charge from electron excess. Like charges repel; unlike charges attract. Charge is conserved (cannot be created or destroyed) and quantised (exists in multiples of the elementary charge e = 1.6 × 10⁻¹⁹ C).
⚡ Electrostatics Core Formulae
Coulomb's Law:
F = k q₁q₂/r² k = 1/(4πε₀) = 9 × 10⁹ N·m²/C²
ε₀ = 8.85 × 10⁻¹² C²/N·m² (permittivity of free space)
Electric Field (E):
E = F/q = kQ/r² (field due to point charge Q at distance r)
Unit: N/C = V/m Dimension: MLT⁻³A⁻¹
Electric Potential (V):
V = kQ/r (potential due to point charge)
Unit: Volt (V) = J/C
Relationship: E = −dV/dr (field = −potential gradient)
Uniform field: V = E × d (d = distance between plates)
Capacitance:
C = Q/V (unit: Farad F = C/V)
Parallel plate: C = ε₀A/d
Quantisation of charge: q = ne (n = integer, e = 1.6 × 10⁻¹⁹ C)
Conservation of charge: total charge in isolated system is constant
The dielectric constant (relative permittivity) κ of a medium reduces the Coulomb force: F = kq₁q₂/(κr²). In vacuum/air: κ = 1. Water: κ ≈ 80 (Coulomb force 80× weaker in water than in vacuum).
Q2. The electric field inside a charged hollow conductor is:
(a) Equal to surface field (b) Zero (c) Maximum at centre (d) Directed outward
Answer: (b) Zero
Inside a conductor in electrostatic equilibrium, free electrons arrange themselves so the net electric field inside is zero. All excess charge resides on the outer surface. This is the principle of electrostatic shielding (Faraday cage). Sensitive electronics use metal enclosures for this reason.
2. Current Electricity
2.1
Electric Current, Ohm's Law & Resistance
The flow of charge — and what resists it
Electric current is the rate of flow of charge. Conventional current flows from positive to negative terminal (outside battery), while electrons flow in the opposite direction. The relationship between current, voltage, and resistance is given by Ohm's Law.
⚡ Current, Resistance & Ohm's Law
Electric Current: I = Q/t Unit: Ampere (A) = C/s
Potential Difference: V = W/Q Unit: Volt (V) = J/C
Resistance: R = V/I Unit: Ohm (Ω)
Ohm's Law: V = IR (valid for ohmic conductors)
V–I graph: straight line through origin → Ohmic conductor
Non-ohmic: diode, filament lamp (R changes with temperature)
Resistivity (ρ):
R = ρL/A
ρ = resistivity of material Unit: Ω·m Dimension: ML³T⁻³A⁻²
L = length of conductor; A = cross-sectional area
Factors affecting resistance:
R ↑ with L (longer wire → more resistance)
R ↓ with A (thicker wire → less resistance)
R ↑ with temperature for metals (ρ increases)
R ↓ with temperature for semiconductors and electrolytes
Conductance (G) = 1/R Unit: Siemens (S)
Conductivity (σ) = 1/ρ Unit: S/m
EMF of battery (ε): ε = V + Ir (terminal voltage = EMF − voltage drop)
r = internal resistance of battery
Short circuit: V = 0, I = ε/r (maximum current)
Resistivity is a material property (like density or thermal conductivity) — it does not depend on the shape or size of the conductor. Resistance depends on both the material (ρ) and the geometry (L, A).
🔌 Resistivity Values (approx.)
Silver: 1.6×10⁻⁸ Ω·m (best conductor)
Copper: 1.7×10⁻⁸ Ω·m (most used)
Aluminium: 2.8×10⁻⁸ Ω·m
Nichrome: 1×10⁻⁶ Ω·m (heating element)
Glass: ~10¹² Ω·m (very high — insulator)
Silicon: ~10³ Ω·m (semiconductor)
🔸 Temperature Effect on Resistance
Metals: R increases with T (positive temp coefficient)
Semiconductors: R decreases with T (negative temp coefficient)
Superconductors: R = 0 at very low temperature
R_T = R₀(1 + αT) — α = temperature coefficient
Nichrome and manganin: R nearly constant with T (used in standards)
📝 TOPIC-WISE PYQ
Current Electricity & Resistance — NDA Pattern Questions
Q1. A wire of resistance 8 Ω is stretched to double its length. Its new resistance is:
(a) 4 Ω (b) 16 Ω (c) 32 Ω (d) 2 Ω
Answer: (c) 32 Ω
Volume = AL is constant when wire is stretched. New L' = 2L, so new A' = A/2. R' = ρL'/A' = ρ(2L)/(A/2) = 4ρL/A = 4R = 32 Ω. Doubling length quadruples resistance (length doubles, area halves → both increase R).
Q2. The resistance of a conductor increases with temperature because:
(a) Electrons move faster (b) More electron-ion collisions (higher lattice vibration) (c) Fewer electrons (d) Voltage increases
Answer: (b) More electron-ion collisions (higher lattice vibration)
As temperature increases, metal ions vibrate more vigorously — this increases the frequency of collisions between drifting electrons and ions, reducing the mean free path and thus increasing resistivity (and resistance). This is why filament bulbs glow — the hot tungsten filament has high resistance.
Q3. A battery of EMF 12 V and internal resistance 2 Ω is connected to a 10 Ω resistor. The terminal voltage is:
(a) 12 V (b) 10 V (c) 2 V (d) 8 V
Answer: (b) 10 V
I = ε/(R+r) = 12/12 = 1 A. Terminal voltage V = ε − Ir = 12 − 1×2 = 10 V. (Or: V = IR_external = 1×10 = 10 V.) The 2 V is dropped across internal resistance.
🤔 TRICKY QUESTIONS
Current Electricity — Reasoning Traps
T1. Two wires A and B of the same material have lengths in ratio 1:2 and diameters in ratio 2:1. What is the ratio of their resistances?
T2. Why is the tungsten filament in a bulb very thin and coiled, while the connecting wires are thick?
Filament: thin (high R → high heat); Wires: thick (low R → no heat loss).
The filament must have very high resistance to generate sufficient heat and light when current flows (P = I²R → high R = more power dissipated as heat). Thin wire → small cross-section A → R = ρL/A is large. The coiling increases effective length in a small volume. Connecting wires must have low resistance to avoid wasting energy — so they are thick (large A → small R). This is the design principle behind all heating elements and filament lamps.
3. Circuits: Series, Parallel & Power
3.1
Series & Parallel Combinations of Resistors
The two fundamental circuit configurations — and what changes between them
⚡ Series & Parallel Resistance Formulae
SERIES CONNECTION:
R_eff = R₁ + R₂ + R₃ + ...
Same current through all resistors: I₁ = I₂ = I₃ = I
Voltage divides: V = V₁ + V₂ + V₃
V_n = I × R_n (voltage proportional to resistance)
R_eff > any individual R (always larger)
PARALLEL CONNECTION:
1/R_eff = 1/R₁ + 1/R₂ + 1/R₃ + ...
Same voltage across all: V₁ = V₂ = V₃ = V
Current divides: I = I₁ + I₂ + I₃
I_n = V/R_n (current inversely proportional to resistance)
R_eff < any individual R (always smaller)
For 2 resistors in parallel: R_eff = R₁R₂/(R₁+R₂)
Kirchhoff's Laws:
KCL (Junction rule): ΣI_in = ΣI_out (charge conservation)
KVL (Loop rule): ΣV = 0 around any closed loop (energy conservation)
In a house, all appliances are connected in parallel — each gets full supply voltage (220–240 V), and failure of one doesn't affect others. Festoon lights (old Christmas lights) used series — one bulb failing broke the whole string.
Fig. 1 — Series circuit (left): same current through all resistors; voltage divides. Parallel circuit (right): same voltage across all; current divides. Household wiring uses parallel — each appliance gets full supply voltage independently.
3.2
Joule's Law of Heating & Electrical Power
Current flowing through resistance generates heat — the basis of all electrical heating
⚡ Joule's Law & Power Formulae
Joule's Law of Heating:
H = I²Rt (heat produced in time t)
H = VIt = V²t/R = I²Rt (all three forms — use as needed)
Unit: Joule (J); 1 kWh = 3.6 × 10⁶ J
Electrical Power:
P = VI = I²R = V²/R Unit: Watt (W) = J/s
1 kW = 1000 W; 1 MW = 10⁶ W
Electrical energy:
E = Pt Unit: kWh (kilowatt-hour) — the "unit" of electricity
1 unit = 1 kWh = energy consumed by 1000 W device in 1 hour
Fuse: rated by maximum safe current; made of low-melting alloy (Sn-Pb)
Fuse current I_fuse = √(P_total / V_supply)
MCB (Miniature Circuit Breaker): modern replacement for fuse
Key for appliances rated P watts at V volts:
R_appliance = V²/P (resistance of the appliance)
I_appliance = P/V (current drawn)
Higher wattage bulb = lower resistance. A 100 W bulb has lower resistance than a 60 W bulb (both at 220 V): R = V²/P → lower P means higher R. This surprises many students.
⚠ NDA Exam Trap — Series vs Parallel for household appliances: When identical bulbs are connected in series, they share voltage — each gets less than rated voltage → dim. In parallel, each gets full voltage → normal brightness. If one bulb fuses in series, all go out. In parallel, only that one goes out. This is why houses use parallel wiring.
Worked Example — Units of Electricity
A 1500 W geyser runs for 2 hours. How many units of electricity does it consume? (Cost = ₹8 per unit)
Energy = Power × Time = 1500 W × 2 h = 3000 Wh = 3 kWh = 3 units
Cost = 3 × ₹8 = ₹24
📝 TOPIC-WISE PYQ
Circuits, Heating & Power — NDA Pattern Questions
Q1. Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:
(a) 11 Ω (b) 1 Ω (c) 3 Ω (d) 2 Ω
Answer: (b) 1 Ω
1/R_eff = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. R_eff = 1 Ω. In parallel, equivalent resistance is always less than the smallest individual resistor (smallest here = 2 Ω).
Q2. A 60 W bulb and a 100 W bulb are connected in series to a 240 V supply. Which bulb glows brighter?
(a) 100 W bulb (b) 60 W bulb (c) Both equally (d) Neither glows
Answer: (b) 60 W bulb glows brighter
R = V²/P. R_60 = 240²/60 = 960 Ω; R_100 = 240²/100 = 576 Ω. In series, same current I flows; P = I²R. Higher R → more power dissipated → brighter. R_60 > R_100, so 60 W bulb glows brighter in series. (In parallel it would be the reverse.)
Q3. An electric iron of 1000 W is used for 30 minutes. The heat generated is:
Answer: (b) 1.8×10⁶ J
H = Pt = 1000 × 30 × 60 = 1000 × 1800 = 1.8×10⁶ J. Always convert time to seconds for joules. (Alternatively: 0.5 kWh = 0.5 × 3.6×10⁶ = 1.8×10⁶ J ✓)
🤔 TRICKY QUESTIONS
Circuits & Power — Classic Exam Traps
T1. A 40 W and a 60 W bulb are rated for 220 V. Which has higher resistance? Which will fuse first if connected in series to 220 V?
40 W has higher resistance; 40 W bulb fuses first in series.
R = V²/P: R_40 = 220²/40 = 1210 Ω; R_60 = 220²/60 = 807 Ω. So 40 W has higher R. In series, same current flows. P = I²R — higher R means more power dissipated. The 40 W bulb receives more power than its rating in this series circuit → it is more likely to overheat and fuse first.
T2. When more appliances are switched on in a house, why does the voltage across each drop slightly?
Increased total current → larger voltage drop across supply line resistance.
The supply line has small resistance r (wires and transformers). Total current I increases when more appliances are added (they are in parallel). Terminal voltage V = V_supply − I×r. As I increases, the drop (I×r) increases, reducing V delivered to appliances. This is why lights dim when a motor or heater is switched on — momentarily high current causes voltage drop.
4. Magnetic Effects of Current & Electromagnetic Induction
4.1
Magnetic Field Due to Current-Carrying Conductors
Oersted's discovery — electric current creates a magnetic field
In 1820, Hans Christian Oersted discovered that a current-carrying wire deflects a compass needle — proving that electric current produces a magnetic field. This is the foundation of all electric motors, generators, and electromagnetic devices used in defence technology.
⚡ Magnetic Field Due to Current
Straight Wire (at distance r):
B = μ₀I / (2πr) (infinite straight wire)
Direction: Right-Hand Thumb Rule (RHR)
Thumb → current direction; Curled fingers → field direction (circles)
Circular Loop (at centre):
B = μ₀I / (2R) (R = radius of loop)
Direction: RHR — curl fingers in direction of current; thumb → field
Solenoid (inside):
B = μ₀nI (n = turns/length)
Direction: RHR for solenoid — thumb → N pole
Key Relations:
B ∝ I (more current → stronger field)
B ∝ 1/r (straight wire: closer → stronger field)
B ∝ n (solenoid: more turns/length → stronger)
Force on current in magnetic field (Motor principle):
F = BIL sinθ (B = field, I = current, L = length, θ = angle)
F = BIL when I ⊥ B (maximum force)
F = 0 when I ∥ B (no force)
Direction: Fleming's Left-Hand Rule (FLHR)
Lorentz force on charge q moving at velocity v:
F = qvB sinθ → F = q(v × B)
Fleming's Left-Hand Rule (FLHR) for motors: Stretch left hand — forefinger: B field direction; middle finger: current direction; thumb: force/motion direction. Used for electric motors. Fleming's Right-Hand Rule (FRHR) is for generators (induced current direction).
Fig. 2 — Magnetic field due to current. Left: Straight wire — circular field lines (B = μ₀I/2πr). Centre: Circular loop — uniform field through centre (B = μ₀I/2R). Right: Solenoid — uniform field inside, behaves like bar magnet.
4.2
Electric Motor & Electromagnetic Induction
Current creates motion (motor); motion creates current (generator) — two sides of the same coin
⚡ Electric Motor Principle
Force on current-carrying conductor in B field: F = BIL
Direction: Fleming's Left-Hand Rule (FLHR)
DC motor: commutator reverses current every half turn
Torque on coil: τ = BINA (I=current, N=turns, A=area)
Converts electrical energy → mechanical energy
Applications: fans, pumps, drills, electric vehicles
🔋 Electromagnetic Induction
Changing magnetic flux induces EMF (Faraday)
Faraday's law: emf = −N dΦ/dt
Lenz's law: induced current opposes change in flux
Generator: converts mechanical energy → electrical
AC generator: slip rings; DC generator: commutator
Transformer: steps AC voltage up/down using EMI
⚡ EMI, Faraday's Law & Transformer
Magnetic Flux: Φ = BA cosθ Unit: Weber (Wb) = T·m²
Faraday's Law: emf = −N × dΦ/dt
N = number of turns; dΦ/dt = rate of change of flux
Induced emf ∝ rate of change of magnetic flux
Lenz's Law: Induced current direction opposes the change that causes it
(Conservation of energy — opposes = braking effect)
Motional EMF: emf = BvL (conductor of length L moving at v in field B)
Transformer:
V_s/V_p = N_s/N_p (turns ratio = voltage ratio)
V_p × I_p = V_s × I_s (power conserved — ideal transformer)
Step-up: N_s > N_p → V_s > V_p (voltage up, current down)
Step-down: N_s < N_p → V_s < V_p (voltage down, current up)
Self-inductance (L): emf = −L × dI/dt Unit: Henry (H)
Mutual inductance (M): emf₂ = −M × dI₁/dt
Lenz's law is the reason you feel resistance when you push a strong magnet into a copper tube — the induced current creates a field opposing the magnet's entry (braking force). This is the basis of eddy current braking used in high-speed trains.
Fig. 3 — Step-up transformer. AC input creates changing flux in iron core; this induces EMF in secondary. V_s/V_p = N_s/N_p. More secondary turns → higher output voltage (lower current). Transformers work only with AC, not DC.
📝 TOPIC-WISE PYQ
Magnetic Effects & EMI — NDA Pattern Questions
Q1. The direction of induced current in a circuit is determined by:
(a) Faraday's law (b) Lenz's law (c) Ohm's law (d) Ampere's law
Answer: (b) Lenz's law Faraday's law gives the magnitude of induced EMF. Lenz's law gives the direction of induced current — it opposes the change in magnetic flux causing it. Together they fully describe electromagnetic induction.
Q2. A transformer has 100 turns in primary and 1000 turns in secondary. If primary voltage is 220 V (AC), the secondary voltage is:
(a) 22 V (b) 220 V (c) 2200 V (d) 22000 V
Answer: (c) 2200 V
V_s/V_p = N_s/N_p → V_s = 220 × (1000/100) = 2200 V. This is a step-up transformer (N_s > N_p). Power lines use step-up transformers to transmit at high voltage (low current → less I²R loss) and step-down near homes.
Q3. Which rule is used to find the direction of force on a current-carrying conductor in a magnetic field?
(a) Right-hand thumb rule (b) Fleming's left-hand rule (c) Fleming's right-hand rule (d) Lenz's law
Answer: (b) Fleming's left-hand rule
FLHR is for motors — it gives the direction of force/motion (thumb) when current (middle finger) flows in a magnetic field (forefinger). Fleming's right-hand rule is for generators — direction of induced current when a conductor moves in a field.
Q4. Transformers work only with AC and not with DC because:
(a) DC has higher voltage (b) DC does not produce changing magnetic flux (c) DC flows in only one direction (d) AC is safer
Answer: (b) DC does not produce changing magnetic flux
EMI requires changing magnetic flux (Faraday's law: emf = −N dΦ/dt). DC produces a constant flux (dΦ/dt = 0) → zero induced EMF in secondary → no output. AC alternates → flux changes continuously → continuous EMF induction in secondary.
🤔 TRICKY QUESTIONS
Magnetic Effects & EMI — Deep Concepts
T1. Why is electrical energy transmitted over long distances at very high voltage (e.g., 400 kV) rather than at low voltage?
High voltage → low current → less power loss (P_loss = I²R).
For a given power P = VI, higher V means lower I. Power lost in transmission line = I²R_line. If V is doubled, I halves, and I²R loss reduces to 1/4. Transmitting at 400 kV instead of 400 V reduces I by 1000×, reducing line losses by 10⁶×. Step-up transformers raise voltage at the power station; step-down transformers reduce it for safe domestic use. This is why AC (which can use transformers) is universally used for power distribution.
T2. A copper disc is dropped through a strong magnetic field and slows down. Why? No current source is connected.
Eddy currents — induced by changing flux — create a braking force (Lenz's law).
As the copper disc moves through the field, the magnetic flux through it changes. By Faraday's law, an EMF is induced, driving circular (eddy) currents within the disc itself. By Lenz's law, these currents create a magnetic field opposing the disc's motion — acting as a brake. This is used in eddy current braking (high-speed trains, roller coasters, precision weighing machines) and in energy meters. Energy is dissipated as heat in the disc — without any external connection.
⚡ High-Yield Formula Sheet — PN07 Electricity
🔋 Electrostatics
F = kq₁q₂/r² (k = 9×10⁹)
E = kQ/r² = F/q (N/C = V/m)
V = kQ/r (Volt)
C = Q/V (Farad)
q = ne (e = 1.6×10⁻¹⁹ C)
🔌 Current & Resistance
I = Q/t (Ampere)
V = IR (Ohm's law)
R = ρL/A (resistivity)
ε = V + Ir (EMF equation)
G = 1/R (Siemens)
📈 Circuits
Series: R_eff = R₁+R₂+R₃
Parallel: 1/R_eff = 1/R₁+1/R₂+1/R₃
Two parallel: R = R₁R₂/(R₁+R₂)
KCL: ΣI_in = ΣI_out
KVL: ΣV = 0 (closed loop)
🔥 Power & Heating
P = VI = I²R = V²/R (Watt)
H = I²Rt = VIt (Joule)
1 kWh = 3.6×10⁶ J
R_appliance = V²/P
I_appliance = P/V
⚡ Magnetic Effects
Straight wire: B = μ₀I/2πr
Loop centre: B = μ₀I/2R
Solenoid: B = μ₀nI
Motor: F = BIL sinθ
FLHR: motors; FRHR: generators
🔄 EMI & Transformer
Φ = BAcosθ (Wb)
emf = −N dΦ/dt (Faraday)
emf = BvL (motional)
V_s/V_p = N_s/N_p
V_p I_p = V_s I_s (ideal transformer)
📌 Key Dimensions: Current = A; Resistance = ML²T⁻³A⁻²; Resistivity = ML³T⁻³A⁻²; Charge = AT; Power = ML²T⁻³; Magnetic flux = ML²T⁻²A⁻¹; Inductance = ML²T⁻²A⁻².
⚡ Quick Revision Booster — PN07 Electricity
🔋 Electrostatics
F = kq₁q₂/r² (k = 9×10⁹ Nm²/C²)
E inside conductor = 0 (Faraday cage)
Charge on outer surface only
q = ne (quantised: multiples of e)
Charge conserved in isolated system
🔌 Ohm's Law & R
V = IR (Ohm's law)
R = ρL/A (double L → double R)
Stretch wire to 2L: R → 4R (volume fixed)
Metal: R↑ with T; Semiconductor: R↓ with T
Terminal V = EMF − Ir
📈 Circuits
Series: I same; V divides; R_eff = ΣR
Parallel: V same; I divides; 1/R = Σ(1/R)
Houses: parallel (each gets full V)
60W bulb in series glows brighter than 100W (higher R)
Parallel R_eff always less than smallest R
🔥 Power & Energy
P = VI = I²R = V²/R
H = I²Rt (Joule heating)
High W bulb → lower R (R = V²/P)
1 unit = 1 kWh = 3.6×10⁶ J
Fuse/MCB: protects against excess current
⚡ Magnetic Effects
RHR thumb: current → field direction (straight wire)
FLHR: motors (force on current in B)
FRHR: generators (motion → induced current)
F = BIL sinθ (max when I ⊥ B)
Solenoid: B = μ₀nI (uniform inside)
🚨 EMI Traps
Faraday: magnitude of emf; Lenz: direction
Lenz's law: induced current opposes flux change
Transformer only works with AC (not DC)
Step-up: N_s > N_p → V_s > V_p; I_s < I_p
Power loss in lines = I²R (high V, low I → less loss)
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