Every electrical device — from a light bulb to a motor — is ultimately governed by the same fundamental question: how much power does it consume, and where does that energy go? Electrical power is the rate at which a device converts electrical energy into other forms (heat, light, mechanical work). Understanding the three forms of the power formula (P = VI = I²R = V²/R), knowing which resistors in a circuit dissipate more power, and converting between joules and kilowatt-hours are skills tested in both board exams and competitive exams every year. Combined with the temperature dependence of resistance (which explains why a light bulb's resistance changes as it heats up), this topic is immediately practical and conceptually rich.
1. Combination of Resistors
Series
Same current through all; voltages add. R_series > largest individual R.
Parallel
Same voltage across all; currents add. R_parallel < smallest individual R.
2. Electrical Power
Electrical power is the rate of doing electrical work — the energy supplied per second to a circuit element:
SI unit: Watt (W) = J/s = V·A
Which Formula to Use?
| Known quantities | Formula |
|---|
| V and I | P = VI |
| I and R | P = I²R |
| V and R | P = V²/R |
3. Power in Series and Parallel — Key Comparisons
| Combination | Fixed quantity | P ∝ | Dissipates more power |
|---|
| Series | Current I (same) | P = I²R ∝ R | Higher resistance |
| Parallel | Voltage V (same) | P = V²/R ∝ 1/R | Lower resistance |
Worked Example
2Ω and 4Ω connected to 12 V:
Parallel: P₁ = 144/2 = 72 W; P₂ = 144/4 = 36 W → 2Ω dissipates more ✓
Series: I = 12/6 = 2 A; P₁ = 4×2 = 8 W; P₂ = 4×4 = 16 W → 4Ω dissipates more ✓
4. Electrical Energy and Units
Commercial unit: kilowatt-hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J
1 unit of electricity = 1 kWh. This is what electricity meters measure.
Example: 100 W lamp run for 10 hours = 100 × 10 = 1000 Wh = 1 kWh = 1 unit
5. Temperature Coefficient of Resistance
α = temperature coefficient of resistance (per °C or per K).
For metals: α > 0 (e.g., copper α ≈ 4×10⁻³/°C). For semiconductors/insulators: α < 0.
Practical implication: A tungsten filament bulb has much higher resistance when hot (glowing) than when cold. At room temperature R ≈ 30 Ω; at 2700 K (operating), R ≈ 480 Ω.
Practice Questions
Q1 (NEET/Board): A 100 W and a 60 W bulb are connected in series to a 220 V supply. Which bulb glows brighter? Find the power dissipated in each.
R_100 = V²/P = 220²/100 = 484 Ω; R_60 = 220²/60 = 806.7 Ω
Series current I = 220/(484+806.7) = 220/1290.7 = 0.170 A
P_100 = I²R_100 = (0.170)² × 484 = 14.0 W
P_60 = I²R_60 = (0.170)² × 806.7 = 23.3 W
The 60 W bulb glows brighter (higher R in series → more power).
Q2 (Board): An electric iron of 1000 W is used for 2 hours daily for 30 days. Calculate the electrical energy consumed in kWh and cost at ₹6 per unit.
Energy = Power × Time = 1 kW × 2 h/day × 30 days = 60 kWh = 60 units
Cost = 60 × ₹6 = ₹360
Q3 (MCQ): A 6 Ω and 3 Ω resistor are connected in parallel across 12 V. The power dissipated in each is:
A) 6Ω: 24W, 3Ω: 48W B) 6Ω: 48W, 3Ω: 24W C) Both 36W D) 6Ω: 12W, 3Ω: 24W
Answer: A) 6Ω: 24 W, 3Ω: 48 W.
P = V²/R: P_6 = 144/6 = 24 W; P_3 = 144/3 = 48 W. In parallel, lower resistance dissipates more. Total = 72 W. Equivalent R = 2 Ω; P_total = 144/2 = 72 W ✓
