Electric Power, Commercial Energy, and Electricity Bills for CBSE Class 10

When you purchase an electric appliance—such as a $100\text{ W}$ LED bulb, a $1500\text{ W}$ room heater, or an $800\text{ W}$ microwave oven—what do these wattage numbers indicate? And when the monthly municipal electricity bill arrives in your mailbox charging for "150 Units", how does the electrical utility meter measure the energy you consumed?

In CBSE Class 10 Science, Chapter 11 (Electricity) concludes with the practical economics of electrical consumption: Electric Power, the commercial billing unit known as the Kilowatt-Hour (kWh), and the step-by-step calculation of household electricity bills.

What You Will Learn

• Definition, mathematical formulas, and SI unit of Electric Power ($P$)
• The four equivalent formulas for electric power: $P = \frac{W}{t} = VI = I^2R = \frac{V^2}{R}$
• Calculating an appliance's internal resistance from its electrical voltage-power rating
• Why the Joule is too small for practical billing: The Kilowatt-Hour (kWh)
• Proof that $1\text{ kWh} = 3.6 \times 10^6\text{ Joules}$ (CBSE Favorite MCQ)
• Step-by-step protocol for calculating household electricity bills
• Solved CBSE board examination numerical problems and common traps

1. What is Electric Power?

In mechanics, power is the rate of doing work. In electricity:

Formal Definition

Electric power is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit per unit time.

The Four Equivalent Power Formulas:

1. From Fundamental Definition: $\mathbf{P = \frac{\text{Work Done } (W)}{\text{Time } (t)} = \frac{\text{Energy Consumed } (E)}{t}}$
2. In Terms of Voltage and Current: Since $W = V \times Q$ and $I = \frac{Q}{t}$: $\mathbf{P = V \times I}$
3. In Terms of Current and Resistance (Series Applications): Substitute $V = I R$ from Ohm's Law: $\mathbf{P = (IR) \times I = I^2 R}$
4. In Terms of Voltage and Resistance (Parallel & Household Applications): Substitute $I = \frac{V}{R}$ from Ohm's Law: $\mathbf{P = V \times \left(\frac{V}{R}\right) = \frac{V^2}{R}}$

2. The SI Unit of Power: The Watt (W)

• The SI unit of electric power is the Watt (W), named after James Watt.
• Definition of $1\text{ Watt}$:

  <u>One Watt ($1 ext{ W}$) is the power consumed by an electrical appliance that draws a current of 1 Ampere when operated at a potential difference of 1 Volt ($1 ext{ W} = 1 ext{ V} imes 1 ext{ A} = 1 ext{ J/s}$).</u>

• Larger Units of Power:
  • $1\text{ Kilowatt (kW)} = 1000\text{ Watts} = 10^3\text{ W}$
  • $1\text{ Megawatt (MW)} = 1000000\text{ Watts} = 10^6\text{ W}$

3. Decoding Appliance Nameplates: Resistance from Ratings

Every commercial appliance carries a factory specification plate such as "$220\text{ V}, 100\text{ W}$". What does this tell you?

• It means that when operated at the standard household line voltage of $220\text{ V}$, it will consume electrical energy at the rate of $100\text{ Joules per second}$.
• Using $P = \frac{V^2}{R}$, you can immediately calculate the internal resistance ($R$) of the appliance: $\mathbf{R = \frac{V^2}{P}}$ $R = \frac{(220)^2}{100} = \frac{48400}{100} = \mathbf{484\ \Omega}$
• Notice that because household appliances are connected in parallel at constant voltage ($220\text{ V}$): $P \propto \frac{1}{R}$ <u>An appliance with higher wattage has a LOWER resistance! A $100 ext{ W}$ bulb has less resistance than a $40 ext{ W}$ bulb!</u>

4. The Commercial Unit of Energy: Kilowatt-Hour (kWh)

The SI unit of energy is the Joule (J) ($1\text{ Joule} = 1\text{ Watt} \times 1\text{ second}$). However, the Joule is an unimaginably tiny unit for commercial billing. A single $100\text{ W}$ light bulb running for just 10 hours consumes $3,600,000\text{ Joules}$! Writing numbers with millions of digits on monthly utility bills is completely impractical.

To resolve this, electricity providers use the Kilowatt-Hour (kWh), colloquially called "One Unit" of electricity on power bills.

Definition of 1 Kilowatt-Hour (kWh):

One kilowatt-hour is the total electrical energy consumed when an electrical appliance of power rating $1\text{ Kilowatt}$ ($1000\text{ W}$) operates continuously for $1\text{ hour}$.

Converting 1 kWh to Joules (CBSE Mandatory Proof):

$\text{Energy} = \text{Power} \times \text{Time}$ $1\text{ kWh} = 1\text{ Kilowatt} \times 1\text{ Hour}$ $1\text{ kWh} = 1000\text{ Watts} \times 3600\text{ Seconds}$ $1\text{ kWh} = 3,600,000\text{ Watt-seconds (Joules)}$

$\mathbf{1\text{ kWh} = 3.6 \times 10^6\text{ Joules}}$

5. Protocol for Household Electricity Bill Calculations

To calculate the monthly electricity cost for a home:

$\mathbf{\text{Electrical Energy in kWh (Units)} = \frac{\text{Total Power (Watts)} \times \text{Operating Hours per Day} \times \text{Number of Days}}{1000}}$ $\mathbf{\text{Total Cost (₹)} = \text{Total Units (kWh)} \times \text{Rate per Unit (₹/kWh)}}$

6. Solved CBSE Board Examination Problems

Solved Example 1: Comparing Bulb Power at Reduced Voltage (CBSE Classic)

Problem: An electric bulb is rated $220\text{ V}$ and $100\text{ W}$. When it is operated on $110\text{ V}$, what power will be consumed?

Solution:

1. Find the Resistance of the Bulb Filament: The physical resistance of the tungsten filament does not change when the applied voltage drops: $R = \frac{V^2}{P} = \frac{220 \times 220}{100} = \mathbf{484\ \Omega}$
2. Calculate New Power at $110\text{ V}$: $P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{110 \times 110}{484}$ $P_{\text{new}} = \frac{12100}{484} = \mathbf{25\text{ W}}$
3. Therefore, <u>the power consumed at $110\text{ V}$ is $25\text{ Watts}$</u>. (Notice: When voltage is halved, power drops to one-fourth ($1/2^2 = 1/4$)!)

Solved Example 2: Complete Household Electricity Bill (NCERT Classic)

Problem: An electric refrigerator rated $400\text{ W}$ operates 8 hours/day. Two electric fans rated $80\text{ W}$ each operate 12 hours/day. What is the cost of energy to operate them for 30 days at ₹$3.00\text{ per kWh}$?

Solution:

1. Energy Consumed by the Refrigerator in 30 Days: $E_{\text{fridge}} = \frac{400\text{ W} \times 8\text{ hours/day} \times 30\text{ days}}{1000} = \frac{96000}{1000} = \mathbf{96\text{ kWh}}$
2. Energy Consumed by the Two Fans in 30 Days:
   • Total power of 2 fans $= 2 \times 80 = 160\text{ W}$. $E_{\text{fans}} = \frac{160\text{ W} \times 12\text{ hours/day} \times 30\text{ days}}{1000} = \frac{57600}{1000} = \mathbf{57.6\text{ kWh}}$
3. Calculate Total Energy Consumed: $E_{\text{total}} = 96 + 57.6 = \mathbf{153.6\text{ kWh (Units)}}$
4. Calculate Total Bill Cost at ₹$3.00$ per kWh: $\text{Cost} = 153.6 \times 3.00 = \mathbf{₹\,460.80}$
5. Therefore, <u>the total cost of electricity for the month is ₹$460.80$</u>.

7. Summary and Examination Tips

Exam Tip: In bulb problems, students often mistakenly think that if voltage drops from $220\text{ V}$ to $110\text{ V}$ (halved), power is halved to $50\text{ W}$. This is wrong! Because power is proportional to the square of voltage ($P \propto V^2$), halving voltage cuts power to one-fourth ($25\text{ W}$)!

Common Mistake: Forgetting to divide by $1000$ when calculating kWh. If power is given in Watts, you MUST divide by $1000$ to convert Watt-hours into Kilowatt-hours (kWh) before multiplying by the cost rate!