Ohm's law, resistance, series & parallel circuits, heating effect, and power
Electric charge (Q): Property of matter that causes force between particles. SI unit: Coulomb (C)
Electric current (I): Rate of flow of electric charge. I = Q/t
SI unit: Ampere (A) = Coulombs/second
Conventional current: Flows from + to – (positive terminal to negative) — opposite to electron flow
Electric potential: Work done to bring unit positive charge from infinity to that point.
Potential difference (V): Work done to move unit charge between two points.
V = W/Q | SI unit: Volt (V)
1 Volt = 1 Joule per Coulomb (1 V = 1 J/C)
At constant temperature, the current through a conductor is directly proportional to the potential difference across its ends.
V ∝ I → V = IR
where R = resistance of the conductor
SI unit of resistance: Ohm (Ω) | 1Ω = 1 V/A
Electricity is like water flowing through a pipe:
• Voltage (V) = water pressure (higher pressure = more flow)
• Current (I) = rate of water flow
• Resistance (R) = narrowness of pipe (more narrow = more resistance = less flow)
R = ρl/A
where ρ = resistivity (or specific resistance), l = length, A = cross-sectional area
• R ∝ l (longer wire → more resistance)
• R ∝ 1/A (thicker wire → less resistance)
• ρ depends on material and temperature
Q: A wire of resistivity 1.6 × 10⁻⁸ Ωm has length 1 m and area 1 mm² = 1 × 10⁻⁶ m². Find resistance.
R = ρl/A = (1.6 × 10⁻⁸ × 1) / (1 × 10⁻⁶) = 1.6 × 10⁻² Ω = 0.016 Ω
Total Resistance: R = R₁ + R₂ + R₃
Same current flows through all components: I = I₁ = I₂ = I₃
Voltage divides: V = V₁ + V₂ + V₃
Disadvantage: If one component fails, all stop working (old Christmas lights!)
1/R = 1/R₁ + 1/R₂ + 1/R₃
Same voltage across all: V = V₁ = V₂ = V₃
Current divides: I = I₁ + I₂ + I₃
Advantage: If one fails, others keep working. Used in home wiring!
Q: Resistors 3Ω, 6Ω are in parallel. The combination is in series with 4Ω. Total voltage = 10V. Find current through each resistor.
Parallel combination: 1/R = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2 → R_parallel = 2Ω
Total R = 4 + 2 = 6Ω
Total I = V/R = 10/6 = 5/3 A
Voltage across 4Ω = (5/3) × 4 = 20/3 V
Voltage across parallel = 10 – 20/3 = 10/3 V
I through 3Ω = (10/3)/3 = 10/9 A | I through 6Ω = (10/3)/6 = 5/9 A
Heat produced in a conductor: H = I²Rt
where H = heat (Joules), I = current (A), R = resistance (Ω), t = time (s)
Also: P = I²R = V²/R = VI (power dissipated as heat)
• Electric bulb: Tungsten filament gets so hot (3000°C) it glows
• Electric iron / heater / toaster: Heating by high-resistance element
• Fuse: Thin wire melts when excess current flows — saves circuit
• Electric kettle: Immersion heater boils water
P = VI = I²R = V²/R
SI unit: Watt (W) = Joules/second
1 kWh (kilowatt-hour) = commercial unit of electrical energy
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ
Electricity bill = Power (kW) × Time (h) × Rate (₹/kWh)
Q: A house uses a 1500W AC, 5 bulbs of 60W each, and a 100W TV for 5 hours daily. Find monthly bill at ₹5/kWh.
Daily power = 1500 + 5×60 + 100 = 1500 + 300 + 100 = 1900 W = 1.9 kW
Daily units = 1.9 × 5 = 9.5 kWh
Monthly units = 9.5 × 30 = 285 kWh
Bill = 285 × 5 = ₹1425
I = Q/t
V = W/Q
V = IR (Ohm's Law)
R = ρl/A
Series: R = R₁+R₂+R₃
Parallel: 1/R = 1/R₁+1/R₂+1/R₃
P = VI = I²R = V²/R
H = I²Rt (Joule's law)
1 kWh = 3.6×10⁶ J