Electric Power MCQs Quiz | Class 10
This quiz is designed for Class X, Subject: Science (Code 086), Unit: Unit IV: Effects of Current, focusing on Electric Power MCQs. It covers the definition of electric power, the fundamental formula P=VI, and the essential relations among P, V, I, and R (P=I²R, P=V²/R). Test your understanding of these core concepts. Submit your answers and download a detailed PDF answer sheet for review.
Understanding Electric Power
Electric power is a fundamental concept in physics and electrical engineering, especially when studying the effects of current. It defines the rate at which electrical energy is consumed or converted into other forms of energy (like heat or light) in an electrical circuit. Understanding electric power is crucial for analyzing circuits, designing electrical systems, and daily applications such as choosing the right appliances.
1. Definition of Electric Power
Electric power is defined as the rate at which electric work is done or electric energy is consumed in an electric circuit. In simpler terms, it tells us how fast energy is being used or supplied. The SI unit of electric power is the Watt (W).
- Watt (W): One watt is defined as the power consumed when one joule of energy is consumed in one second (1 W = 1 J/s).
- Kilowatt (kW): A larger unit, 1 kW = 1000 W.
2. Formulas for Electric Power (P=VI and its relations)
The relationship between electric power (P), potential difference (V), current (I), and resistance (R) are derived using Ohm’s Law and the basic definition of power.
a) Power in terms of Voltage and Current (P = VI)
This is the most fundamental formula for electric power. If a current (I) flows through a conductor across which a potential difference (V) is maintained, the power consumed is given by:
P = V × I
- P: Power (in Watts)
- V: Potential difference (in Volts)
- I: Current (in Amperes)
b) Power in terms of Current and Resistance (P = I²R)
Using Ohm’s Law (V = IR), we can substitute V into the P = VI formula:
P = (IR) × I = I²R
- This formula is particularly useful when the current and resistance of a component are known, and it highlights the heating effect of current.
c) Power in terms of Voltage and Resistance (P = V²/R)
Again using Ohm’s Law (I = V/R), we can substitute I into the P = VI formula:
P = V × (V/R) = V²/R
- This formula is useful when the voltage across a component and its resistance are known. It shows that for a constant voltage, power is inversely proportional to resistance.
Summary of Electric Power Formulas
| Formula | Variables Involved | When to Use |
|---|---|---|
| P = VI | Power, Voltage, Current | Most general; if V and I are known. |
| P = I²R | Power, Current, Resistance | When current and resistance are known; highlights heating effect. |
| P = V²/R | Power, Voltage, Resistance | When voltage and resistance are known; for constant voltage applications. |
3. Commercial Unit of Electric Energy: Kilowatt-hour (kWh)
While power is the rate of energy consumption, electrical energy consumed over a period is measured commercially in kilowatt-hour (kWh). This is what electricity meters in homes measure.
- 1 kilowatt-hour (kWh) = 1 unit of electricity.
- Conversion to Joules: 1 kWh = 1000 W × 3600 s = 3,600,000 J or 3.6 × 106 J.
Quick Revision Points:
- Electric power is the rate of energy consumption (P = W/t).
- The SI unit of power is Watt (W).
- Three main formulas: P = VI, P = I²R, P = V²/R.
- P = I²R shows the heating effect (Joule’s Law of Heating).
- P = V²/R implies that for constant voltage, lower resistance means higher power.
- Kilowatt-hour (kWh) is the commercial unit of electrical energy.
- 1 kWh = 3.6 × 106 Joules.
Extra Practice Questions:
- A current of 5 A flows through a bulb connected to a 220 V supply. Calculate the power of the bulb.
- An electric heater of resistance 20 ohms draws a current of 10 A from the mains. Calculate the power consumed by the heater.
- An electric lamp is rated 100 W, 220 V. What is the resistance of the lamp?
- How much energy in joules is consumed by a 60 W bulb in 5 minutes?
- If two bulbs, one 60 W and one 100 W, are connected in parallel to the same voltage supply, which bulb will draw more current and why?
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