Prerequisites
Before studying this topic, make sure you understand:
- Ohm’s Law - V-I relationship
- Resistance - Energy dissipation mechanism
- Work and Energy - Basic energy concepts
The Hook: Why Does Your Phone Charger Get Warm?
You’re binge-watching Stranger Things on your phone. After 2 hours of continuous charging, you touch the charger - it’s WARM! Why?
The answer: Electrical energy is being WASTED as heat in the charger and cable. This is inevitable whenever current flows through resistance.
More examples:
- Electric kettles boil water using this heating effect
- Incandescent bulbs waste 90% of energy as heat (only 10% as light!)
- Iron Man’s arc reactor would overheat instantly without sci-fi cooling
- Your laptop fan works overtime because CPU resistors generate heat
The Big Question: How much electrical energy is consumed per second? And how much becomes heat? Let’s calculate!
Interactive Demo
Adjust voltage and resistance to see power dissipation:
The Core Concept: Electrical Power
Power is the rate at which electrical energy is consumed or delivered.
$$\boxed{P = \frac{W}{t} = \frac{E}{t}}$$Unit: Watt (W) = Joule/second
In simple terms: “How many joules of electrical energy are used per second”
Power in Terms of V and I
From work done in moving charge:
$$W = VQ$$ $$P = \frac{W}{t} = \frac{VQ}{t} = V \times \frac{Q}{t}$$ $$\boxed{P = VI}$$This is the MASTER formula for electrical power!
Multiple Forms of Power Formula
Using Ohm’s law ($V = IR$), we can derive different forms:
1. Basic Form
$$\boxed{P = VI}$$Use when: V and I are directly given
2. Current Form
$$\boxed{P = I^2R}$$Derivation: $P = VI = (IR) \times I = I^2R$
Use when: Current and resistance are known
3. Voltage Form
$$\boxed{P = \frac{V^2}{R}}$$Derivation: $P = VI = V \times \frac{V}{R} = \frac{V^2}{R}$
Use when: Voltage and resistance are known
| Given | Use Formula | Example |
|---|---|---|
| V and I | $P = VI$ | Power consumed by any device |
| I and R | $P = I^2R$ | Heat in resistor, same I |
| V and R | $P = V^2/R$ | Devices in parallel, same V |
Common mistake: Using wrong formula based on what’s given!
JEE Trick: In series (same I) → Use $P = I^2R$ In parallel (same V) → Use $P = V^2/R$
Joule’s Law of Heating
When current flows through a resistor, electrical energy is converted to heat.
Heat produced:
$$\boxed{H = I^2Rt = Pt}$$where:
- $H$ = heat energy (Joules)
- $I$ = current (A)
- $R$ = resistance (Ω)
- $t$ = time (seconds)
Alternative forms:
$$H = VIt = \frac{V^2t}{R}$$In calories:
$$H = \frac{I^2Rt}{4.2} \text{ calories}$$(Since 1 calorie = 4.2 J)
An electric heater rated 1000 W running for 1 hour:
Energy consumed: $E = Pt = 1000 \times 3600 = 3.6 \times 10^6$ J = 3.6 MJ
This equals: 1 kWh (kilowatt-hour) = 1 “unit” of electricity!
Electricity bill: You pay for ENERGY (kWh), not power (W)!
Applications of Heating Effect
1. Electric Heater / Kettle
Mechanism: Nichrome wire (high resistance) → Current → Heat → Boils water
Why Nichrome?
- High resistivity → More heat per unit length
- High melting point (1400°C) → Won’t melt
- Low temperature coefficient → Constant performance
2. Electric Fuse
Purpose: Protects circuits from excessive current
Mechanism:
- Thin wire with low melting point (tin/lead alloy)
- If $I$ exceeds safe limit → $I^2R$ heating melts the wire
- Circuit breaks → Prevents fire
Design: Rated for specific current (5A, 10A, 15A fuses)
Using thick wire or aluminum foil instead of proper fuse is EXTREMELY DANGEROUS!
What happens:
- Thick wire won’t melt even at high current
- Excessive current → House wiring heats up
- Can cause FIRE!
Always use correct rated fuse.
3. Electric Bulb (Incandescent)
Mechanism:
- Tungsten filament at ~2500°C glows white-hot
- Power input: 100 W
- Light output: ~10 W (10% efficiency!)
- Heat waste: ~90 W (90% wasted as heat)
Why tungsten?
- Highest melting point of all metals (3422°C)
- Can glow white-hot without melting
This is why: LEDs are better - 90% efficiency, almost no heat!
4. Electric Iron
Temperature control:
- Bimetallic strip acts as thermostat
- When too hot → Strip bends → Breaks circuit
- When cool → Strip straightens → Completes circuit
5. Welding
High current (100-400 A) through contact point:
- $H = I^2Rt$ with large $I$ → Intense heat
- Metals melt and fuse together
Power Rating of Appliances
Household appliances are marked with power ratings.
Example: “220V, 100W” bulb
Meaning:
- Designed to work at 220 V
- Consumes 100 W at this voltage
- Current drawn: $I = P/V = 100/220 = 0.45$ A
- Resistance: $R = V^2/P = 48400/100 = 484$ Ω
What if Voltage Changes?
Case 1: Bulb rated “220V, 100W” connected to 110V
Resistance stays constant: $R = \frac{V^2}{P} = \frac{220^2}{100} = 484$ Ω
At 110V:
$$P' = \frac{V'^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25 \text{ W}$$Result: Bulb glows dimly (only 25W instead of 100W)
Case 2: Same bulb connected to 440V
$$P' = \frac{440^2}{484} = \frac{193600}{484} = 400 \text{ W}$$Result: Bulb draws 4 times rated power → Will burn out instantly!
Lower voltage than rated:
- Device works but at reduced power (dim light, slow heating)
- Generally safe
Higher voltage than rated:
- Excessive power → Overheating
- Can damage or destroy device
- FIRE HAZARD!
Always check voltage ratings before using appliances!
Comparing Bulbs: Which Glows Brighter?
In Series Connection
Same current through both bulbs.
Power: $P = I^2R$
Higher resistance → Higher power → Brighter glow
Example: 100W and 60W bulbs (both rated 220V) in series
$$R_{100} = \frac{220^2}{100} = 484 \text{ Ω}$$ $$R_{60} = \frac{220^2}{60} = 806.7 \text{ Ω}$$Since $R_{60} > R_{100}$:
In series: 60W bulb glows BRIGHTER! (counterintuitive!)
In Parallel Connection
Same voltage across both bulbs.
Power: $P = \frac{V^2}{R}$
Lower resistance → Higher power → Brighter glow
Since $R_{100} < R_{60}$:
In parallel: 100W bulb glows BRIGHTER! (as expected)
Series: “Same I, so P = I²R → High R wins”
- 60W bulb (higher R) glows brighter
Parallel: “Same V, so P = V²/R → Low R wins”
- 100W bulb (lower R) glows brighter
Mnemonic: “Series likes Stubborn (high R), Parallel likes Powerful (low R, high P rating)”
Maximum Power Transfer Theorem
For a source with internal resistance $r$ connected to external load $R$:
Power delivered to load:
$$P = I^2R = \left(\frac{E}{R+r}\right)^2 R$$Maximum power when:
$$\boxed{R = r}$$Maximum power:
$$P_{max} = \frac{E^2}{4r}$$Efficiency at maximum power: 50% (half energy wasted in internal resistance!)
Audio systems: Speaker impedance matched to amplifier output Power transmission: NOT used (efficiency more important than max power) Battery charging: Fast charging needs impedance matching
JEE problems often test: Finding R for maximum power, or calculating efficiency
Electrical Energy and Commercial Unit
Energy
$$E = Pt = VIt = I^2Rt = \frac{V^2t}{R}$$SI Unit: Joule (J)
Commercial Unit: Kilowatt-hour (kWh) or “Unit”
$$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$$Calculating Electricity Bill
Example: 5 bulbs of 100W each, used 6 hours daily for 30 days
Energy consumed:
$$E = 5 \times 100 \text{ W} \times 6 \times 30 \text{ hours} = 90,000 \text{ Wh} = 90 \text{ kWh}$$If cost is Rs. 5 per unit:
$$\text{Bill} = 90 \times 5 = \text{Rs. } 450$$Memory Tricks & Patterns
Mnemonic for Power Formulas
“Power Very Impressive” → $P = VI$
“Powerful Indians Ride Reliably” → $P = I^2R$
“Vinod’s Victory Realizes” → $P = V^2/R$
Series vs Parallel Power
“Series → Same I → I² matters → High R wins”
“Parallel → Same V → V²/R matters → Low R wins”
Unit Conversion
“Kill a Watt, Hire 3600 joules” → 1 kWh = 3.6 × 10⁶ J
JEE Pattern Recognition
- If “in series” mentioned: Use $P = I^2R$
- If “in parallel” mentioned: Use $P = V^2/R$
- If “brighter bulb” asked: Calculate power dissipated
- If “electricity bill” asked: Find energy in kWh
When to Use This
Power calculation:
- Given V, I → Use $P = VI$
- Given I, R (series) → Use $P = I^2R$
- Given V, R (parallel) → Use $P = V^2/R$
Heating problems:
- Heat produced → Use $H = I^2Rt$
- Temperature rise → Connect $H = mc\Delta T$
Comparing bulbs:
- In series → Compare using $P = I^2R$ (higher R brighter)
- In parallel → Compare using $P = V^2/R$ (lower R brighter)
Efficiency:
- $\eta = \frac{\text{Useful output}}{\text{Input}} \times 100\%$
Common Mistakes to Avoid
Wrong: “Always use P = VI”
Correct: Choose based on what’s CONSTANT in the circuit!
- Series (constant I) → Use $P = I^2R$
- Parallel (constant V) → Use $P = V^2/R$
Example trap: Two resistors 2Ω and 4Ω in series with 6A current. Power in each?
Wrong approach: “Need V for each, use P = VI” Right approach: Same I! Use $P = I^2R$ directly
- $P_1 = 6^2 \times 2 = 72$ W
- $P_2 = 6^2 \times 4 = 144$ W
Wrong: “100W bulb always consumes 100W”
Correct: 100W is power consumed at RATED voltage only!
Example: “100W, 220V” bulb at 110V:
- Rating tells us $R = V^2/P = 484$ Ω (constant)
- At 110V: $P = V^2/R = 110^2/484 = 25$ W (NOT 100W!)
Wrong: “100W bulb is always brighter than 60W bulb”
Correct: In SERIES, it’s OPPOSITE!
Reason:
- $R_{100} = 484$ Ω (from rating)
- $R_{60} = 806.7$ Ω (from rating)
- In series: $P = I^2R$ (same I)
- Higher R → Higher P → 60W bulb glows brighter!
Memory: Ratings are for parallel (home wiring), series behavior is opposite!
Wrong: “H = I²Rt gives heat in watts”
Correct: This gives heat in JOULES!
Units check:
- $I^2Rt$ = A² × Ω × s = W × s = J (energy, not power!)
- For power (W), use $P = I^2R$ (without time)
Example: 2A through 5Ω for 10s
- Heat: $H = 2^2 \times 5 \times 10 = 200$ J
- Power: $P = 2^2 \times 5 = 20$ W
Practice Problems
Level 1: Foundation (NCERT/Basic)
A 100W, 220V bulb is operated at rated voltage. Find (a) current drawn, (b) resistance.
Solution:
(a) Current:
$$P = VI$$ $$I = \frac{P}{V} = \frac{100}{220} = 0.45 \text{ A}$$(b) Resistance:
$$R = \frac{V}{I} = \frac{220}{0.45} = 488.9 \text{ Ω}$$Or directly:
$$R = \frac{V^2}{P} = \frac{220^2}{100} = 484 \text{ Ω}$$Answer: (a) 0.45 A, (b) 484 Ω
An electric heater of resistance 10Ω takes 5A current. Find power consumed and heat produced in 2 minutes.
Solution:
Power:
$$P = I^2R = 5^2 \times 10 = 250 \text{ W}$$Heat in 2 min = 120 s:
$$H = Pt = 250 \times 120 = 30,000 \text{ J} = 30 \text{ kJ}$$Or:
$$H = I^2Rt = 25 \times 10 \times 120 = 30,000 \text{ J}$$Answer: Power = 250 W, Heat = 30 kJ
Calculate electricity bill for a 1000W heater used 2 hours daily for 30 days at Rs. 6 per kWh.
Solution:
Energy consumed:
$$E = Pt = 1000 \text{ W} \times 2 \times 30 = 60,000 \text{ Wh} = 60 \text{ kWh}$$Bill:
$$\text{Cost} = 60 \times 6 = \text{Rs. } 360$$Answer: Rs. 360
Level 2: JEE Main
Two bulbs rated “100W, 220V” and “60W, 220V” are connected in series to 220V supply. Which glows brighter?
Solution:
Step 1: Find resistances from ratings
$$R_{100} = \frac{V^2}{P} = \frac{220^2}{100} = 484 \text{ Ω}$$ $$R_{60} = \frac{V^2}{P} = \frac{220^2}{60} = 806.7 \text{ Ω}$$Step 2: In series, same current $I$
Total: $R_{total} = 484 + 806.7 = 1290.7$ Ω
$$I = \frac{220}{1290.7} = 0.17 \text{ A}$$Step 3: Power dissipated (use $P = I^2R$ since same I)
$$P_{100} = I^2 R_{100} = (0.17)^2 \times 484 = 14 \text{ W}$$ $$P_{60} = I^2 R_{60} = (0.17)^2 \times 806.7 = 23.5 \text{ W}$$Answer: 60W bulb glows BRIGHTER (counterintuitive!)
Quick method: In series, higher R dissipates more power Since $R_{60} > R_{100}$, 60W bulb wins!
A “100W, 220V” bulb is connected to 110V supply. Find power consumed and compare brightness.
Solution:
Resistance (constant):
$$R = \frac{220^2}{100} = 484 \text{ Ω}$$At 110V:
$$P = \frac{V^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25 \text{ W}$$Answer: Power = 25 W (only 25% of rated power!)
Brightness: 1/4 of normal → Much dimmer!
Three 60Ω resistors are connected (a) in series, (b) in parallel across 120V. Find power in each case.
Solution:
(a) Series:
$$R_{eq} = 60 + 60 + 60 = 180 \text{ Ω}$$ $$P = \frac{V^2}{R_{eq}} = \frac{120^2}{180} = \frac{14400}{180} = 80 \text{ W}$$(b) Parallel:
$$\frac{1}{R_{eq}} = \frac{1}{60} + \frac{1}{60} + \frac{1}{60} = \frac{3}{60}$$ $$R_{eq} = 20 \text{ Ω}$$ $$P = \frac{V^2}{R_{eq}} = \frac{120^2}{20} = \frac{14400}{20} = 720 \text{ W}$$Answer: (a) 80 W, (b) 720 W
Insight: Parallel draws 9 times more power! (Lower equivalent R)
Level 3: JEE Advanced
A battery of EMF 12V and internal resistance 2Ω is connected to variable external resistance R. Find R for maximum power transfer and calculate the maximum power.
Solution:
Maximum power condition:
$$R = r = 2 \text{ Ω}$$Current at this condition:
$$I = \frac{E}{R + r} = \frac{12}{2 + 2} = 3 \text{ A}$$Maximum power delivered to R:
$$P_{max} = I^2R = 3^2 \times 2 = 18 \text{ W}$$Or directly:
$$P_{max} = \frac{E^2}{4r} = \frac{144}{4 \times 2} = 18 \text{ W}$$Efficiency:
$$\eta = \frac{R}{R+r} = \frac{2}{4} = 50\%$$Answer: R = 2Ω, P_max = 18W, Efficiency = 50%
Key insight: At maximum power, efficiency is only 50% - half the energy is wasted in internal resistance!
An electric kettle has two coils. When one is used, water boils in 10 min. When the other is used, water boils in 20 min. How long will it take if (a) both coils are used in series, (b) both in parallel?
Solution:
Let the coils have resistances $R_1$ and $R_2$.
Power: $P = \frac{V^2}{R}$
Time: $t \propto R$ (for same heat required)
From given data: $R_1 \propto 10$ min → $R_1 = 10k$ (where k is constant) $R_2 \propto 20$ min → $R_2 = 20k$
(a) Series:
$$R_s = R_1 + R_2 = 10k + 20k = 30k$$ $$t_s = 30 \text{ min}$$(b) Parallel:
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10k} + \frac{1}{20k} = \frac{3}{20k}$$ $$R_p = \frac{20k}{3}$$ $$t_p = \frac{20}{3} = 6.67 \text{ min}$$Answer: (a) 30 min, (b) 6.67 min
Shortcut for similar problems:
- Series: Add times → 10 + 20 = 30 min
- Parallel: Reciprocal addition → $\frac{1}{t_p} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$ → $t_p = 6.67$ min
Two resistors $R_1 = 100$ Ω and $R_2 = 200$ Ω are connected in series to a 300V supply. If heat developed in $R_1$ in time $t_1$ equals heat developed in $R_2$ in time $t_2$, find $t_1/t_2$.
Solution:
In series, same current $I$ flows.
Heat in $R_1$: $H_1 = I^2 R_1 t_1$ Heat in $R_2$: $H_2 = I^2 R_2 t_2$
Given: $H_1 = H_2$
$$I^2 R_1 t_1 = I^2 R_2 t_2$$ $$R_1 t_1 = R_2 t_2$$ $$\frac{t_1}{t_2} = \frac{R_2}{R_1} = \frac{200}{100} = 2$$Answer: $t_1/t_2 = 2$ or $t_1 : t_2 = 2:1$
Key insight: For same heat in series circuit, time is inversely proportional to resistance!
- Smaller R needs MORE time
- Larger R needs LESS time
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Basic power | $P = VI$ |
| Power in resistor (current known) | $P = I^2R$ |
| Power in resistor (voltage known) | $P = V^2/R$ |
| Heat produced | $H = I^2Rt = Pt$ |
| Series connection (bulbs) | Use $P = I^2R$ → Higher R brighter |
| Parallel connection (bulbs) | Use $P = V^2/R$ → Lower R brighter |
| Maximum power transfer | $R = r$, $P_{max} = E^2/(4r)$ |
| Energy in kWh | $E = Pt$ (P in kW, t in hours) |
| Electricity bill | Energy (kWh) × Rate per unit |
JEE Strategy: High-Yield Points
Bulb comparison in series vs parallel (VERY HIGH YIELD!)
- Memorize: Series → Higher R brighter, Parallel → Higher P rating brighter
- Practice 10+ variations
Power formula selection
- Series → $P = I^2R$ (same current)
- Parallel → $P = V^2/R$ (same voltage)
- Don’t blindly use $P = VI$!
Voltage change effect on power
- Find R from rating: $R = V^2/P$
- Then calculate new power at new voltage
- Common in numerical problems
Maximum power transfer
- Condition: $R = r$
- Efficiency: 50% at maximum power
- Often combined with battery problems
Heating effect applications
- Electric kettle time problems
- Fuse wire calculations
- Water boiling problems
Time-saving tricks:
- For bulb ratings, remember: $R = V^2/P$ first, then proceed
- Series bulbs: No need to calculate I, just compare R values
- Parallel bulbs: Compare power ratings directly
- Heating time problems: $t \propto R$ (for same heat, same V)
Related Topics
Within Current Electricity
- Ohm’s Law - Foundation for power calculations
- Resistance - Why power is dissipated as heat
- Kirchhoff’s Laws - Power in complex circuits
- Wheatstone Bridge - Balanced condition means zero power
Connected Chapters
- Work, Energy, Power - Energy concepts
- Thermodynamics - Heat and temperature
- Magnetic Effects - Power in electromagnetic devices
Math Connections
- Quadratic Equations - Maximum power problems
- Proportionality - P ∝ I², P ∝ V²
- Optimization - Finding maximum power conditions
Teacher’s Summary
Three forms of power: $P = VI = I^2R = V^2/R$ - choose based on what’s constant (I in series, V in parallel)!
Heating effect: $H = I^2Rt$ - electrical energy converted to heat, unavoidable in resistors
Bulbs in series: Higher resistance glows brighter (use $P = I^2R$). In parallel: Higher power rating glows brighter (use $P = V^2/R$)
Maximum power transfer when load resistance equals source internal resistance ($R = r$), but efficiency is only 50%
Power rating (100W, 220V) is at RATED voltage only - actual power changes if voltage changes according to $P = V^2/R$ where R is constant
“Electrical power is energy per second, and most of it becomes heat in resistors - that’s why your charger gets warm, your bulb wastes energy, and Iron Man needs a massive cooling system!”