CBSE Class 10 Science Electricity Notes

ELECTRIC CHARGE:

  Symbol: Q    SI Unit: Coulomb (C)

  Charge of one electron: e = −1.6 × 10⁻¹⁹ C

  Charge of one proton:   e = +1.6 × 10⁻¹⁹ C

  Number of electrons in 1 Coulomb:

  n = Q / e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons

  One Coulomb = charge carried by 6.25 × 10¹⁸ electrons

ELECTRIC CURRENT:

  I  =  Q / t

  where:  I = electric current (Ampere, A)

          Q = electric charge (Coulomb, C)

          t = time (second, s)

  SI Unit: Ampere (A)  =  1 Coulomb per second (1 C/s)

  1 milliampere (mA) = 10⁻³ A

  1 microampere (µA) = 10⁻⁶ A

  Direction of conventional current: Positive → Negative (outside battery)

  Direction of electron flow: Negative → Positive (opposite to conventional current)

Conventional Current vs Electron Flow

ELECTRON FLOW: from negative terminal to positive terminal of battery (outside).

CONVENTIONAL CURRENT: from positive terminal to negative terminal (outside battery).

Why opposite? Historical convention — established before electrons were discovered.

Benjamin Franklin assumed positive charges flowed. The direction was already accepted

by the time electrons were discovered by J.J. Thomson (1897).

In all calculations and circuit analysis: use CONVENTIONAL CURRENT direction.

The physics and the maths work out identically either way.

ELECTRIC POTENTIAL DIFFERENCE (Voltage):

  V  =  W / Q

  where:  V = potential difference (Volt, V)

          W = work done in moving charge from one point to another (Joule, J)

          Q = charge moved (Coulomb, C)

  SI Unit: Volt (V)  =  1 Joule per Coulomb (1 J/C)

  Definition: 1 Volt = potential difference between two points if 1 Joule of work

  is done in moving 1 Coulomb of charge from one point to the other.

  Potential difference is also called VOLTAGE.

Aspect

EMF (Electromotive Force)

Terminal Voltage

Definition

Total energy supplied by cell per unit charge pushed through the complete circuit

Potential difference across the terminals of the cell when current is flowing

Formula

EMF = W_total / Q

V_terminal = EMF − I × r

When equal?

EMF = Terminal voltage only when no current flows (open circuit)

Terminal voltage < EMF when current flows (due to internal resistance)

Symbol

ε (epsilon) or E

V

Measured by

Voltmeter across cell in open circuit

Voltmeter across cell in closed circuit

OHM'S LAW:

  'The current through a conductor is directly proportional to the potential

   difference across its ends, provided physical conditions

   (temperature, material) remain constant.'

  V  ∝  I   (at constant temperature)

  V  =  I × R

  R  =  V / I

  I  =  V / R

  where:  V = potential difference / voltage (Volt, V)

          I = current (Ampere, A)

          R = resistance (Ohm, Ω)

  SI Unit of Resistance: Ohm (Ω)

  1 Ohm = 1 Volt per Ampere (1 V/A)

V-I Graph for Ohmic Conductors

For a conductor obeying Ohm's Law (an 'ohmic conductor'):

  • V-I graph is a STRAIGHT LINE passing through the origin.

  • Slope of V-I graph = R (resistance) — steeper slope = higher resistance.

  • I-V graph is also a straight line; slope = 1/R (conductance).

Non-ohmic conductors (e.g., diodes, filament bulbs, thermistors):

  • V-I graph is NOT a straight line → resistance changes with current/voltage.

  • A filament bulb: resistance INCREASES as it heats up (R increases with temp).

  • A diode: allows current in one direction only → asymmetric I-V curve.

Worked Numerical 3.1 — Ohm's Law

Problem: A wire carries a current of 2 A when connected to a 12 V battery.

Find: (a) resistance of wire, (b) current if voltage increased to 18 V.

Given: I = 2 A, V = 12 V

(a) Using R = V/I:

    R = 12 / 2 = 6 Ω

(b) At 18 V, using I = V/R:

    I = 18 / 6 = 3 A

Note: Since temperature is assumed constant, R remains 6 Ω.

RESISTANCE FORMULA:

  R  =  ρ × l / A

  where:  R = resistance (Ω)

          ρ = resistivity (also called specific resistance) (Ω·m)

          l = length of conductor (m)

          A = cross-sectional area of conductor (m²)

  RELATIONSHIPS:

  R ∝ l        (longer wire → more resistance — more collisions)

  R ∝ 1/A      (thicker wire → less resistance — more paths for electrons)

  R depends on ρ (material property — different for different materials)

  R increases with temperature (for metals — more ionic vibration → more collisions)

Material

Resistivity (Ω·m) at 20°C

Category

Silver (Ag)

1.60 × 10⁻⁸

Conductor (best conductor)

Copper (Cu)

1.69 × 10⁻⁸

Conductor (most used in wiring)

Aluminium (Al)

2.82 × 10⁻⁸

Conductor (used in overhead lines)

Tungsten (W)

5.60 × 10⁻⁸

Conductor (used in light bulb filaments)

Nickel-Chrome (Nichrome)

1.10 × 10⁻⁶

Alloy conductor (used in heating elements)

Carbon (graphite)

3.5 × 10⁻⁵

Semiconductor

Silicon (Si)

6.40 × 10²

Semiconductor

Glass

10¹⁰ − 10¹⁴

Insulator

Rubber

10¹³ − 10¹⁶

Insulator

Why Tungsten is Used in Bulb Filaments

Tungsten has a VERY HIGH melting point (3422°C) — can withstand extremely high temperatures.

Tungsten has HIGH resistivity — generates sufficient heat/light at reasonable currents.

Tungsten is ductile — can be drawn into very thin wires.

Why Nichrome is Used in Heating Elements (electric iron, room heater, toaster):

Nichrome has HIGH resistivity — produces more heat per unit length.

Nichrome has HIGH melting point — can withstand high operating temperatures.

Nichrome does not oxidise at high temperatures — longer lifespan.

Worked Numerical 4.1 — Resistance and Resistivity

Problem: A copper wire of length 2 m and cross-sectional area 0.5 mm² carries

a current. Find its resistance. (ρ_copper = 1.69 × 10⁻⁸ Ω·m)

Given: l = 2 m,  A = 0.5 mm² = 0.5 × 10⁻⁶ m²,  ρ = 1.69 × 10⁻⁸ Ω·m

Using R = ρl/A:

  R = (1.69 × 10⁻⁸ × 2) / (0.5 × 10⁻⁶)

  R = (3.38 × 10⁻⁸) / (5.0 × 10⁻⁷)

  R = 0.0676 Ω  ≈  0.068 Ω

The copper wire has a very low resistance — expected for a conductor.

SERIES CIRCUIT RULES:

  1. CURRENT: Same current flows through every component

     I_total = I₁ = I₂ = I₃ = ... = I_n

  2. VOLTAGE: Total voltage = sum of voltages across each component

     V_total = V₁ + V₂ + V₃ + ... + V_n

     (voltage DIVIDES across components)

  3. EQUIVALENT RESISTANCE: Sum of all individual resistances

     R_s = R₁ + R₂ + R₃ + ... + R_n

  Key facts:

  → R_s is ALWAYS GREATER than any individual resistance

  → If one component fails (open circuit), ALL components stop working

  → Voltage across each resistor: V_i = I × R_i  (proportional to R_i)

Worked Numerical 5.1 — Series Circuit

Problem: Three resistors R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 10 Ω are connected in series

to a 20 V battery. Find: (a) equivalent resistance, (b) current through circuit,

(c) voltage across each resistor.

Given: R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 10 Ω, V = 20 V

(a) Equivalent resistance:

    R_s = R₁ + R₂ + R₃ = 4 + 6 + 10 = 20 Ω

(b) Current through circuit:

    I = V / R_s = 20 / 20 = 1 A

    (same 1 A flows through every resistor)

(c) Voltage across each resistor:

    V₁ = I × R₁ = 1 × 4 = 4 V

    V₂ = I × R₂ = 1 × 6 = 6 V

    V₃ = I × R₃ = 1 × 10 = 10 V

    Check: V₁ + V₂ + V₃ = 4 + 6 + 10 = 20 V ✓ (equals battery voltage)

PARALLEL CIRCUIT RULES:

  1. VOLTAGE: Same voltage across every branch (component)

     V_total = V₁ = V₂ = V₃ = ... = V_n

  2. CURRENT: Total current = sum of currents through each branch

     I_total = I₁ + I₂ + I₃ + ... + I_n

     (current DIVIDES across branches)

  3. EQUIVALENT RESISTANCE:

     1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R_n

  For TWO resistors only (special case):

     R_p = (R₁ × R₂) / (R₁ + R₂)  [Product over Sum]

  Key facts:

  → R_p is ALWAYS LESS than the smallest individual resistance

  → If one branch fails, other branches continue to work

  → Current through each branch: I_i = V / R_i  (inversely proportional to R_i)

Worked Numerical 6.1 — Parallel Circuit

Problem: Three resistors R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω are connected in parallel

to a 12 V battery. Find: (a) equivalent resistance, (b) total current,

(c) current through each resistor.

Given: R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω, V = 12 V

(a) Equivalent resistance:

    1/R_p = 1/6 + 1/12 + 1/4

    1/R_p = 2/12 + 1/12 + 3/12 = 6/12 = 1/2

    R_p = 2 Ω  (less than smallest R₃ = 4 Ω ✓)

(b) Total current:

    I_total = V / R_p = 12 / 2 = 6 A

(c) Current through each resistor (same 12 V across all):

    I₁ = 12 / 6  = 2 A

    I₂ = 12 / 12 = 1 A

    I₃ = 12 / 4  = 3 A

    Check: I₁ + I₂ + I₃ = 2 + 1 + 3 = 6 A = I_total ✓

Feature

Series Circuit

Parallel Circuit

Connection

Resistors joined end-to-end — one single path

Resistors joined side-by-side — multiple paths

Current

SAME through all components (I₁ = I₂ = I₃)

DIFFERENT in each branch; sum = I_total

Voltage

DIVIDES across components (V₁ + V₂ = V_total)

SAME across all branches (V₁ = V₂ = V_total)

Equivalent Resistance

R_s = R₁ + R₂ + R₃ (sum)

1/R_p = 1/R₁ + 1/R₂ + 1/R₃

R_eq compared to R_i

GREATER than any individual R

LESS than smallest individual R

Effect of failure

One failure stops ALL components

One failure does not affect other branches

Current distribution

No distribution — same current everywhere

Inversely proportional to resistance of branch

Voltage distribution

Proportional to resistance (V_i = I × R_i)

No distribution — same voltage everywhere

Household use?

NO — each appliance would get different V

YES — every appliance gets same mains voltage

Battery drain

Less total current drawn (higher R)

More total current drawn (lower R_p)

Example use

Resistor chains to limit current; series bulbs (old Christmas lights)

All home appliances; industrial loads

Worked Numerical 7.1 — Mixed Series-Parallel Circuit

Problem: R₁ = 3 Ω is in series with a parallel combination of R₂ = 6 Ω and R₃ = 6 Ω.

Total battery voltage = 12 V. Find: (a) total equivalent resistance, (b) total current,

(c) voltage across R₁, (d) voltage across parallel combination, (e) current through R₂ and R₃.

Step 1: Reduce the parallel combination:

  R₂₃ = (R₂ × R₃)/(R₂ + R₃) = (6 × 6)/(6 + 6) = 36/12 = 3 Ω

Step 2: Now circuit is R₁ (3 Ω) in series with R₂₃ (3 Ω):

  R_total = R₁ + R₂₃ = 3 + 3 = 6 Ω

Step 3: Total current:

  I = V / R_total = 12 / 6 = 2 A

  (This 2 A flows through R₁ and then splits into the parallel branches)

Step 4: Voltage across R₁:

  V₁ = I × R₁ = 2 × 3 = 6 V

Step 5: Voltage across parallel combination:

  V₂₃ = I × R₂₃ = 2 × 3 = 6 V  (or: V₂₃ = 12 − 6 = 6 V ✓)

Step 6: Current through each parallel branch (6 V across each):

  I₂ = V₂₃ / R₂ = 6 / 6 = 1 A

  I₃ = V₂₃ / R₃ = 6 / 6 = 1 A

  Check: I₂ + I₃ = 1 + 1 = 2 A = I_total ✓

JOULE'S LAW OF HEATING:

  H  =  I² × R × t

  where:  H = heat produced (Joule, J)

          I = current (Ampere, A)

          R = resistance (Ohm, Ω)

          t = time (second, s)

  Derivation from W = QV and Q = It and V = IR:

  W = QV = (It)(V) = It(IR) = I²Rt

  Also:   H  =  V × I × t   =   VIt       (using V = IR → V²t/R)

          H  =  V²t / R

  Heat is proportional to: I² (square of current), R, and t

Understanding Joule's Law — H ∝ I²

Why H ∝ I²? Doubling the current QUADRUPLES the heat produced.

This is why high currents are dangerous — heat increases very rapidly with current.

In a fuse: when current exceeds safe limit, I² heating melts the fuse wire → circuit breaks.

In a heater: high resistance + high current → large H generated → room warms up.

In a bulb: high resistance tungsten filament + current → very high T → produces light.

Key point: heat produced per second = POWER dissipated = I²R watts.

Worked Numerical 8.1 — Joule's Law

Problem: A 40 W electric bulb is connected to 220 V mains.

Find: (a) current through bulb, (b) resistance of filament, (c) heat produced in 1 hour.

Given: P = 40 W, V = 220 V, t = 1 hour = 3600 s

(a) Current:

    P = VI → I = P/V = 40/220 = 2/11 A ≈ 0.182 A

(b) Resistance:

    R = V/I = 220/(2/11) = 220 × 11/2 = 1210 Ω

    OR: R = V²/P = 220²/40 = 48400/40 = 1210 Ω ✓

(c) Heat produced in 1 hour:

    H = I²Rt = (2/11)² × 1210 × 3600

    H = (4/121) × 1210 × 3600 = 40 × 3600 = 144000 J = 144 kJ

    OR: H = Pt = 40 × 3600 = 144000 J = 144 kJ ✓

ELECTRIC POWER — THREE FORMS:

  P  =  V × I          (power = voltage × current)

  P  =  I² × R        (using V = IR to eliminate V)

  P  =  V² / R        (using I = V/R to eliminate I)

  where:  P = power (Watt, W)

          V = voltage (V), I = current (A), R = resistance (Ω)

  SI Unit: Watt (W)  =  1 Joule per second (1 J/s)

  1 kilowatt (kW) = 1000 W

  ELECTRICAL ENERGY CONSUMED:

  E  =  P × t  =  V × I × t  (in Joules)

  COMMERCIAL UNIT OF ELECTRICAL ENERGY:

  1 kilowatt-hour (kWh) = 1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 10⁶ J

  Also called 1 unit of electricity (what you pay for on your electric bill).

Kilowatt-Hour (kWh) — The Unit on Your Electricity Bill

1 kWh = energy consumed by a 1000 W (1 kW) appliance in 1 hour.

How to calculate electricity bill:

  1. Find power of each appliance in kW (divide Watts by 1000)

  2. Multiply by hours of use → energy in kWh

  3. Multiply by tariff (cost per kWh) → cost in rupees

Example: A 2000 W heater used for 3 hours:

  Energy = 2 kW × 3 h = 6 kWh

  If tariff = ₹5 per unit: Cost = 6 × 5 = ₹30

1 kWh = 3.6 × 10⁶ J = 3,600,000 J  (important conversion to remember)

Worked Numerical 9.1 — Electric Power and Energy

Problem: An electric iron draws 5 A at 220 V. How much electrical energy does it

consume in 2 hours? Express your answer in (a) Joules and (b) kWh.

Given: I = 5 A, V = 220 V, t = 2 h = 7200 s

Power: P = VI = 220 × 5 = 1100 W = 1.1 kW

(a) Energy in Joules:

    E = P × t = 1100 × 7200 = 7,920,000 J = 7.92 × 10⁶ J

(b) Energy in kWh:

    E = P(kW) × t(hours) = 1.1 × 2 = 2.2 kWh

This iron consumes 2.2 units of electricity in 2 hours.

Mistake

Why It Is Wrong

Correct Understanding

Ammeter connected in parallel

Ammeter has very low resistance — parallel connection would short-circuit the component

Ammeter MUST be in SERIES. Voltmeter MUST be in PARALLEL.

R_parallel = R₁ + R₂

That is the series formula

Parallel: 1/R_p = 1/R₁ + 1/R₂. Series: R_s = R₁ + R₂.

R_parallel is larger than individual R

Adding parallel paths reduces resistance

R_parallel is ALWAYS LESS than smallest individual resistor.

H = IRt (missing the square)

The square on I is critical — common error

Joule's Law: H = I²Rt. Forgetting the square gives completely wrong answer.

Calculating kWh using seconds

kWh requires hours, not seconds

E(kWh) = P(kW) × t(hours). Or E(J) = P(W) × t(seconds).

Fuse connected in neutral wire

Fuse in neutral doesn't protect against live wire faults

Fuse must always be in the LIVE wire — that carries the dangerous high voltage.

Conventional current flows from − to +

This is electron flow direction

Conventional current flows from + to − (outside battery). Electrons flow − to +.

R = ρA/l (A and l swapped)

Resistance increases with length, decreases with area

R = ρl/A. Longer wire → more R. Thicker wire (larger A) → less R.

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

Slope depends on which axis is which

V-I graph (V on y-axis): slope = R. I-V graph (I on y-axis): slope = 1/R.

Term / Formula

Definition / Value

Electric Charge (Q)

Q in Coulombs (C). Charge of electron = −1.6 × 10⁻¹⁹ C. 1 C = 6.25 × 10¹⁸ electrons.

Electric Current (I)

I = Q/t. Rate of flow of charge. SI unit: Ampere (A) = 1 C/s.

Potential Difference (V)

V = W/Q. Work done per unit charge. SI unit: Volt (V) = 1 J/C.

EMF

Total energy per unit charge supplied by a source; equals terminal voltage in open circuit.

Ohm's Law

V = IR (at constant temperature). Resistance R = V/I. SI unit: Ohm (Ω) = 1 V/A.

Resistance (R)

Opposition to current flow. R = ρl/A. Increases with temperature in metals.

Resistivity (ρ)

Material property. ρ = RA/l. Unit: Ω·m. Independent of dimensions.

Series Equivalent R

R_s = R₁ + R₂ + R₃. Always greater than any individual R.

Parallel Equivalent R

1/R_p = 1/R₁ + 1/R₂. Always less than smallest individual R.

Two resistors in parallel

R_p = (R₁ × R₂)/(R₁ + R₂)  [Product over Sum]

Joule's Law of Heating

H = I²Rt = VIt = V²t/R. Unit: Joule (J).

Electric Power (P)

P = VI = I²R = V²/R. SI unit: Watt (W) = 1 J/s.

Electrical Energy (E)

E = Pt = VIt. In Joules (J) or kWh.

1 kWh

1 kilowatt-hour = 1000 W × 3600 s = 3.6 × 10⁶ J = 1 unit of electricity

Fuse

Low melting point wire in series; melts when current exceeds rated value; protects circuit

Earthing

Connecting metal casing of appliance to Earth; safety measure; prevents electric shock

MCB

Miniature Circuit Breaker; electromagnetic switch; trips on excess current; can be reset

Ammeter

Measures current; connected in SERIES; ideal resistance = 0

Voltmeter

Measures voltage; connected in PARALLEL; ideal resistance = ∞

Short Circuit

Live and neutral wires in contact; huge current; fuse blows; fire hazard

Overloading

Too many appliances on one circuit; current > safe limit; wire overheats; fire hazard

Board Exam Strategy for Electricity

1. ALL numericals: always write R = V/I or P = VI form first, then rearrange — avoid jumping to rearranged formula directly.

2. Series vs Parallel: the most commonly confused topic. R_series = sum. 1/R_parallel = sum of reciprocals.

3. R_parallel is ALWAYS LESS than smallest R. R_series is ALWAYS GREATER than largest R. Use as check.

4. Joule's Law: H = I²Rt — never forget the SQUARE on I. This is the #1 formula error in this chapter.

5. Power formulas: P = VI = I²R = V²/R — know all three; use whichever two quantities you have.

6. kWh: use P in kW and t in hours. 1 kWh = 3.6 × 10⁶ J (very frequently asked conversion).

7. For mixed circuits: simplify parallel groups first, then add series. Verify current/voltage checks.

8. Ammeter: series, low R. Voltmeter: parallel, high R. Fuse: live wire only, series, low melting point.

9. V-I graph: straight line through origin = ohmic conductor. Slope = R. Non-linear = non-ohmic.

10. For 5-mark circuit questions: draw circuit, find R_eq step by step, find I, find V across each part.