Electrical Energy, Power and Efficiency

Question 1

1. From the definition of potential difference, the electrical work done is given by the equation
   
   Where
   W = work
   Q = charge
   V = potential difference
2. Since the work done must be equal to the energy to do the work, therefore we can also say that, the electrical energy ( E ) is also given by the formula

Question 2

Given that the potential difference across a bulb is 240V and the current that flow through the bulb is 0.25A. Find the energy dissipated in the bulb in 30s.

Answer 2

Formula of current,

Question 3

Power

1. The electric power, P is defined as the rates of energy that supply to the circuit ( or the rates of work been done ) by sources of electric.
2. The unit of electric power is the watt (W).
3. One watt of power equals the work done in one second by one volt of potential difference in moving one coulomb of charge.
4. The electrical power of an electric circuit component can be find from the following equations:
   
   Where
   P = power
   t = time
   I = current
   V = potential difference
   R = resistance

Question 4

A current of 0.50A flows through a 100Ω resistor. What is the power lost in the resistor?

Question 5

An electric iron has a heating element of resistance 50Ω. If the operating current flowing through it is 4A, calculate the heat energy produced in 2 minutes.

Answer 5

Power of the iron,

Question 6

What is the power dissipated in a 4Ω light bulb connected to a 12V battery? What is the power dissipated in a 2Ω light bulb connected to the same battery? Which bulb is brighter?

Answer 6

Assume that the bulbs are resistor

Power dissipated in the 4Ω resistor,

Power dissipated in the 2Ω resistor,

The power of the 2Ω bulb is higher, hence it is brighter.

[Conclusion: The lower the load resistance in a circuit, the greater the power dissipated in the circuit]

Question 7

The figure above shows that an ideal battery is connected in parallel to two resistors with resistances 2Ω and 4Ω. Find the power dissipated in

a. the 4Ω resistor
b. the 2Ω resistor

Answer 7

a. The potential difference across the 2 resistor = 12V
The power of the 2Ω resistor,

b.
The power of the 4Ω resistor,

[Conclusion: In a parallel connection, the lower the resistance, the greater the power of the resistor.]

Question 8

An ideal battery with e.m.f. 12 V is connected in series to two bulbs with resistances R₁ = 4Ω and R₂ = 2Ω What is the current in the circuit and the power dissipation in each bulb?

Answer 8

Potential difference across the 2 resistors, V = 12V
Equivalence resistance of the 2 resistors, R = 4 + 2 = 6Ω
Current in the circuit,

Power dissipated in R₁

Power dissipated in R₂

[Conclusion: In a series connection, the greater the resistance of a resistor, the greater the power dissipated]

Question 9

Sum of the Power

In a circuit of any connection (series or parallel), the power dissipated in the whole circuit is equal to the sum of the power dissipated in each of the individual resistor.

Example:
2 identical bulb of resistance 3Ω is connected to an e.m.f. of 12V. Find the power dissipated in the circuit if
a. the bulb is connected in series
b. if the bulb is connected in parallel

Answer:
a.

Current pass through the 2 resistors,

Power of each of the resistor,

Sum of the power,

b.

Potential difference across the 2 resistor = 12V
Power of each of the resistor,

Sum of the power,

Question 10

A 800W heater is used to heat 250 cm³ of water from 30 to 100°C. What is the minimum time in which this can be done? [Density of water = 1000kg/m³; Specific Heat Capacity of water = 4200J°C^-1 kg^-1]

Answer 10

Energy supply by the heater, E = Pt

Heat energy absorbed by the water, E = mcθ

Let’s assume that all the energy supplied by the heater is converted to heat energy and absorbed by the water, hence

Question 11

Power Rating

1. Figure above is an example of power rating label.
2. An electrical appliance which is marked 240V, 1200W means that the electrical appliance will consume 1200J of energy in every second if the potential difference across it is 240V.

Question 12

A bulb rated 240V/80W is operated from a 120V power source. Find the resistance and the current flows through it.

Answer 12

The current flowing through the bulb

Question 13

Energy Consumption

Calculating the cost of electricity consumption

1. The amount of electrical energy consumed in a given time:
   
2. The larger the power rating in the electrical appliance, the higher energy is used for every second.
3. The longer the usage time, the higher electrical energy is consumed.
4. The cost of electricity consumption is based on the number of kilowatt-hours (kWh) of electrical energy used.
5. The kilowatt-hours are sometimes known as the domestic units of electricity.
6. The kilowatt-hour (kWh) is the energy used by a device at a rate of 1000 watts in one hour.
   1 kWh = (1000 W) × (60 × 60 s) = 3.6 MJ

Question 14

If TNB charges 22 cents for each kWh of electrical energy used, calculate the total cost of using a 2kW electric kettle for 15 minutes and a 20 W filament bulb for 8 hours.

Answer 14

Electrical energy consumed by the kettle,

Electrical energy consumed by the bulb,

Total energy consumed,

Question 15

Efficiency of Electrical Appliance

1. The efficiency of an electrical appliance is given by the following equation

2. Normally, the efficiency of an electrical appliance is less than 100% due to the energy lost as heat and the work done against friction in a machine.

Question 16

A lamp is marked “240V, 50W”. If it produces a light output of 40W, what is the efficiency of the lamp?

Question 17

An electric motor raises a mass of 2kg to a height of 5m in 10s. If the input current from a source of 12V is 2A, find the efficiency of the electric motor.

Answer 17

Input power,

Output power

Question 18

Steps to Save Electricity

1. Use efficient lighting
2. Buy efficient electric appliances.
3. Use appliances with automatic power off function.
4. Choose electrical appliances of sizes and features which best suit your needs.
5. Proper utilization of all electrical appliances
   
   1. Defrost refrigerators regularly
   2. Run your washing machine only when it is fully loaded & Iron your clothes only when you have at least a few pieces to iron.
   3. Regular cleaning of air filters in air-condition units and clothes dryers.