Electricity

Why This Chapter Matters

10-12 marks in CBSE board exams. Ohm's law, series/parallel resistors, power, and numerical problems tested every year. Heavily numerical - practice calculations!

Prerequisites

•Charge, current, circuit diagrams from Class 7-8

•Basic algebra for solving equations

Core Concepts

1. Electric Current and Potential Difference

Current (I): I = Q/t

Q = charge (Coulombs), t = time (seconds)

Unit: Ampere (A) = C/s

Ammeter: connected in SERIES

Potential Difference (V): V = W/Q

W = work done (Joules), Q = charge (Coulombs)

Unit: Volt (V) = J/C

Voltmeter: connected in PARALLEL

2. Ohm's Law

V = IR (at constant temperature)

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

Graph of V vs I: straight line through origin.

Resistance (R):

R = rho x L/A (rho = resistivity, L = length, A = cross-sectional area)

R increases with: length, temperature (for metals)

R decreases with: cross-sectional area

3. Combination of Resistors

Series: R_total = R1 + R2 + R3

•Same current through each

•Voltage divides proportionally

Parallel: 1/R_total = 1/R1 + 1/R2 + 1/R3

•Same voltage across each

•Current divides inversely proportional to resistance

For two in parallel: R = (R1 x R2)/(R1 + R2)

Parallel resistance < smallest individual resistance (always!)

Why household appliances in parallel?

•Each gets full voltage

•If one fails, others work

•Can switch independently

4. Electric Power

P = VI = I^2 x R = V^2/R

Unit: Watt (W) = J/s

Commercial unit: kilowatt-hour (kWh) = 1000W for 1 hour = 3.6 x 10^6 J

1 kWh = 1 unit on electricity bill

5. Heating Effect (Joule's Law)

H = I^2 x R x t

Applications: electric bulb, iron, toaster, heater, fuse

Fuse: Low melting point wire. Melts at excess current. Protects appliances.

Why tungsten for bulb filament? Very high melting point (3380 C), high resistivity, low evaporation rate.

Solved Examples

Q1: R = 10 ohm, I = 0.5 A. Find V.

V = IR = 0.5 x 10 = 5V

Q2: Three resistors 2, 3, 6 ohm in parallel. Find R_total.

1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1

R = 1 ohm

Q3: 100W, 220V bulb used 2 hours. Energy consumed?

Energy = 100W x 2h = 200 Wh = 0.2 kWh = 0.2 units

Q4: Electric iron, R = 20 ohm, I = 3A, t = 30s. Heat produced?

H = I^2 x R x t = 9 x 20 x 30 = 5400 J

PYQs

2023: Define resistance. SI unit?

Resistance = opposition to flow of current. SI unit = Ohm (1 ohm = 1 V/A).

2022: R1=5, R2=10, R3=20 in series, V=20V. Find current.

R_total = 35 ohm. I = V/R = 20/35 = 4/7 A ≈ 0.57 A

2021: Units consumed by 60W bulb in 10 hours?

Energy = 60 x 10 = 600 Wh = 0.6 kWh = 0.6 units. Cost at Rs5/unit = Rs3.

2020: Why is tungsten used in electric bulb filament?

High melting point (3380 C) - survives high temperature.

High resistivity - generates enough heat to emit light.

MCQ Practice

Q1. Ohm's law valid when:

(A) High temperature (B) Temperature changes (C) Temperature constant (D) Low current

Answer: C

Q2. In parallel combination, same across all:

(A) Current (B) Resistance (C) Power (D) Voltage

Answer: D

Q3 (Hard). Wire of R stretched to double length. New resistance?

Answer: 4R (length doubles: R x 2; area halves: R x 2 more; total: 4R)

Revision Notes

FORMULAS:
I = Q/t (current = charge/time)
V = W/Q (potential difference = work/charge)
V = IR (Ohm's Law)

SERIES: R = R1 + R2 + R3 (current same, voltage divides)
PARALLEL: 1/R = 1/R1 + 1/R2 + 1/R3 (voltage same, current divides)
Two in parallel: R = R1xR2/(R1+R2)

POWER: P = VI = I^2R = V^2/R (Watts)
ENERGY: E = Pt (Joules) | 1 kWh = 3.6 x 10^6 J

HEATING: H = I^2 x R x t (Joules)

INSTRUMENTS:
Ammeter: SERIES (measures current)
Voltmeter: PARALLEL (measures voltage)

Common mistakes: Ammeter in parallel (will short circuit!). Parallel R is NOT bigger than individual R (it's always smaller).