Electrical Energy and Power Dissipation: Formula, Explanation & Solved Problems

19.9 ENERGY AND ELECTRICAL POWER

19.9.1 Electrical Energy

Suppose that we connect an electrical energy supply (e.g. battery, dynamo, solar cell) to a metallic conductor. Immediately, an electric field will be set up within the conductor.

When an electric field is set up within a metallic conductor, the conduction electrons will experience electric forces acting on them. They accelerate and gain kinetic energy. On the other hand, the field loses electric potential energy.

Since the electrons move within the crystal lattice structure of the conductor, they frequently interact with the lattices. These interactions cause the electrons to slow down and consequently lose kinetic energy. This process is continuous as the electrons drift through the conductor.

As the conduction electrons interact with the lattice, energy is transferred to the lattice points. The amplitude of lattice vibration increases, resulting in an increase in temperature of the metallic conductor.

Conclusion: Electrical energy is transformed into thermal energy when current flows in a metallic conductor.

19.9.2 Electrical Power Dissipation

Suppose a potential difference V is applied across two points on a conductor, producing a current I. If the amount of charge transferred in time interval Δt is Δq, then the work done is:

$\boxed{W = \Delta qV}$

The energy supplied by the electrical source in time Δt is also W. Hence, the electrical power P is:

$\boxed{P = \frac{W}{\Delta t} = \left(\frac{\Delta q}{\Delta t}\right)V}$

Since $I = \Delta q/\Delta t$, we obtain:

$\boxed{P = VI}$

Alternative Forms of Power Equation

Using Ohm’s Law:

$V = IR$

We get:

$\boxed{P = I(IR) = I^2R}$

Also, since:

$I = V/R$

We obtain:

$\boxed{P = (V/R)V = \frac{V^2}{R}}$

Solved Example of Electrical Power Dissipation in a Heater (Step-by-Step Solution)

Example 19.13

Figure 19.11 Electric hot water heater diagram

An electric heater of resistance 5.0 Ω is immersed in a beaker containing 0.50 kg of water at 25.0°C. A 24 V DC supply is connected to the heater for 5.0 minutes.

(a) Determine the electrical power generated by the heater.
(b) Determine the current in the heater.
(c) Determine the temperature of the water after 5.0 minutes. State two assumptions made.

Specific heat capacity of water = 4.2 kJ kg^-1 K^-1

Jawab:

Given:

• Resistance, $R = 5.0 \Omega$
• Mass of water, $m = 0.50 \ kg$
• Initial temperature, $T_i= 25.0 \ ^0C$
• Voltage, $V = 24 \ V$
• Time, $t = 5.0 \ minutes = 300 \ s$
• Specific heat capacity, $c = 4.2 \times 10^3 \ J kg^{-1}K^{-1}$

(a) Electrical Power

We use the formula:

$P = V^2/ R$

$P = 24^2/5 = 576/5 = 115.2 \ W$

Answer: The electrical power generated is 115.2 W.

(b) Current in the heater

We use Ohm’s Law:

$I = V/R$

$I = 24/5 = 4.8 A

Answer: The current is 4.8 A.

(c) Final temperature of the water

Step 1: Calculate electrical energy supplied

$E = P \times t$

$E = 115.2 \times 300 = 34,560 \ J$

Step 2: Use heat equation

$Q = mc \Delta T$

$34,560 = 0.50 \times (4.2 \times 10^3) \times \Delta T$

$34,560 = 2100 \times \Delta T$

$\Delta T = 34,560/2100 = 16.46 \ ^0C$

Step 3: Final temperature

$T_{final} = T_{initial} + \Delta T$

$T_{final} = 25.0 + 16.46 = 41.5 \ ^0C$ (approx)

Answer: The final temperature of the water is approximately 41.5°C.

Assumptions:

• No heat is lost to the surroundings (perfect insulation).
• All electrical energy is converted into heat energy in the water.

Example 19.14 Integrated Concepts

Figure 19.12 A home vaporizer that produces water vapor by passing an electric current through water. Submerged electrodes conduct the current, causing the water to evaporate and release vapor from the top of the device.

A home vaporizer produces water vapor by passing an electric current through water, causing it to evaporate with only a slight increase in temperature. The device operates at a current of 4.00 A and is connected to a 110 V AC power supply with an efficiency of 90.0%.

(a) Determine the rate at which water is vaporized (in grams per minute).

(b) Calculate the total mass of water required for 7.0 hours of continuous operation.

Jawab

Given:

• Current, I = 4.00 A
• Voltage, V = 110 V
• Efficiency, η = 90.0% = 0.90
• Time, t = 7.0 h = 7 × 3600 = 25,200 s
• Latent heat of vaporization of water, L = 2.26 × 10⁶ J/kg

Concept:

Electrical power is given by:

P = IV

Only a fraction of this power is used to vaporize water due to efficiency:

P_useful = η × IV

The energy required to convert water into vapor is:

Q = mL

Thus, the rate of vaporization is:

m/t = P_useful / L

(a) Vaporization rate

Step 1: Calculate electrical power

P = IV = 4.00 × 110 = 440 W

Step 2: Calculate useful power

P_useful = 0.90 × 440 = 396 W

Step 3: Calculate mass rate

m/t = P_useful / L

m/t = 396 / (2.26 × 10⁶) = 1.75 × 10^-4 kg/s

Convert to grams per minute:

m/t = 1.75 × 10^-4 × 1000 × 60 = 10.5 g/min

Answer (a): The vaporization rate is 10.5 g/min.

(b) Total mass of water

Step 1: Use mass rate

m/t = 1.75 × 10^-4 kg/s

Step 2: Multiply by total time

m = (1.75 × 10^-4) × 25,200 = 4.41 kg

Answer (b): The total mass of water required is 4.41 kg.

Explanation:

The electrical energy supplied to the vaporizer is converted into heat energy. Due to efficiency, only 90% of the electrical power is used to vaporize water. The energy required for vaporization depends on the latent heat, which determines how much energy is needed to convert liquid water into vapor without changing temperature.

Example 19.15 Integrated Concepts

An electric car with a mass of 800 kg is powered by a 12.0 V battery system. The motor operates with an efficiency of 92.0%.

Figure 19.13 A compact electric hatchback is being recharged at a public charging station on a city street.

(a) What current must the battery supply to accelerate the car from rest to 20.0 m/s in 50.0 s?

(b) What current is required for the car to climb a hill of height 150 m in 120 s while moving at a constant speed of 20.0 m/s and overcoming a resistive force of 400 N?

(c) What current is needed to maintain a constant speed of 20.0 m/s while overcoming the same resistive force?

Jawab

Given:

• Mass, m = 800 kg
• Voltage, V = 12.0 V
• Efficiency, η = 0.92
• Final velocity, v = 20.0 m/s
• Time (a), t = 50.0 s
• Height, h = 150 m
• Time (b), t = 120 s
• Resistive force, F = 400 N
• g = 9.8 m/s²

Concept:

Electrical power supplied by the battery is:

P = IV

Useful mechanical power is reduced by efficiency:

P_useful = η × IV

We also use:

• Kinetic energy: KE = ½mv²
• Gravitational potential energy: PE = mgh
• Power: P = Work / time

(a) Acceleration from rest

Step 1: Calculate kinetic energy gained

KE = ½mv² = ½(800)(20²) = 160,000 J

Step 2: Calculate useful power

P_useful = KE / t = 160,000 / 50 = 3,200 W

Step 3: Calculate electrical power

P = P_useful / η = 3,200 / 0.92 = 3,478 W

Step 4: Calculate current

I = P / V = 3,478 / 12 = 290 A (approx)

Answer (a): The required current is 290 A.

(b) Climbing the hill

Step 1: Calculate potential energy

PE = mgh = 800 × 9.8 × 150 = 1,176,000 J

Step 2: Calculate work against resistance

Distance = v × t = 20 × 120 = 2,400 m

W = F × d = 400 × 2,400 = 960,000 J

Step 3: Total work

W_total = 1,176,000 + 960,000 = 2,136,000 J

Step 4: Useful power

P_useful = W / t = 2,136,000 / 120 = 17,800 W

Step 5: Electrical power

P = 17,800 / 0.92 = 19,348 W

Step 6: Current

I = 19,348 / 12 = 1,612 A (approx)

Answer (b): The required current is 1,612 A.

(c) Constant speed motion

Step 1: Power to overcome resistance

P_useful = F × v = 400 × 20 = 8,000 W

Step 2: Electrical power

P = 8,000 / 0.92 = 8,696 W

Step 3: Current

I = 8,696 / 12 = 725 A (approx)

Answer (c): The required current is 725 A.

Explanation:

The battery supplies electrical energy which is converted into mechanical energy by the motor. Due to efficiency losses, more electrical power is required than the useful mechanical power. During acceleration, energy increases kinetic energy. While climbing, energy is used to increase gravitational potential energy and overcome resistive forces. At constant speed, power is only needed to balance resistive forces.

Example 19.16

An electric power station needs to transmit 8.00 × 10⁷ W of electrical power through transmission lines at a voltage of 600 V. The total resistance of the transmission lines is 0.80 Ω.

(a) What current is required to transmit this power?

(b) How much power is dissipated as heat in the transmission lines?

(c) What is unrealistic about this result?

(d) Which assumptions are unreasonable or inconsistent?

Jawab

Given:

• Power, P = 8.00 × 10⁷ W
• Voltage, V = 600 V
• Resistance, R = 0.80 Ω

Concept:

Electrical power is given by:

P = IV

Power loss in transmission lines due to resistance is:

P_loss = I²R

Efficient power transmission requires high voltage and low current to minimize losses.

(a) Current required

Step 1: Use power formula

I = P / V

I = (8.00 × 10⁷) / 600

I = 1.33 × 10⁵ A

Answer (a): The required current is 1.33 × 10⁵ A.

(b) Power dissipated in the lines

Step 1: Use power loss formula

P_loss = I²R

P_loss = (1.33 × 10⁵)² × 0.80

P_loss = (1.77 × 10¹⁰) × 0.80

P_loss = 1.42 × 10¹⁰ W

Answer (b): The power loss is 1.42 × 10¹⁰ W.

(c) What is unrealistic?

The power loss (1.42 × 10¹⁰ W) is much greater than the transmitted power (8.00 × 10⁷ W). This means more energy is lost than delivered, which is clearly impractical.

Answer (c): The result is unrealistic because the energy loss exceeds the transmitted power.

(d) Unreasonable assumptions

• The transmission voltage (600 V) is too low for long-distance power transmission.
• This leads to an extremely large current, causing huge energy losses.
• In real systems, very high voltages (kV range) are used to reduce current and minimize losses.

Answer (d): The assumption of using low voltage for large power transmission is unrealistic and inconsistent with practical systems.

Explanation:

This example highlights why electrical power is transmitted at high voltages. Since power loss depends on the square of current (I²R), reducing current significantly lowers energy loss. Therefore, power companies use transformers to increase voltage and improve efficiency.

Example 19.17

A water heater is used to heat 1200 kg of water from 15°C to 45°C. The system operates with an efficiency of 80.0% due to heat losses to the surroundings. The cost of electricity is $0.12 per kWh.

(a) Calculate the total cost of heating the water.

(b) If the heating process takes 3.0 hours using a 220 V AC heater, determine the current drawn.

Jawab

Given:

• Mass of water, m = 1200 kg
• Initial temperature, T_i = 15°C
• Final temperature, T_f = 45°C
• Temperature change, ΔT = 30°C
• Specific heat capacity of water, c = 4.2 × 10³ J kg^-1 K^-1
• Efficiency, η = 0.80
• Cost of electricity = $0.12/kWh
• Time, t = 3.0 h = 10,800 s
• Voltage, V = 220 V

Concept:

The heat energy required to raise the temperature of water is:

Q = mcΔT

Due to inefficiency, the electrical energy required is:

E = Q / η

Electrical energy in kWh is:

1 kWh = 3.6 × 10⁶ J

Power is given by:

P = E / t and P = IV

(a) Cost of heating

Step 1: Calculate heat energy

Q = mcΔT = 1200 × (4.2 × 10³) × 30

Q = 1.512 × 10⁸ J

Step 2: Account for efficiency

E = Q / η = (1.512 × 10⁸) / 0.80

E = 1.89 × 10⁸ J

Step 3: Convert to kWh

E = (1.89 × 10⁸) / (3.6 × 10⁶) = 52.5 kWh

Step 4: Calculate cost

Cost = 52.5 × 0.12 = $6.30

Answer (a): The cost of heating the water is $6.30.

(b) Current used

Step 1: Calculate power

P = E / t = (1.89 × 10⁸) / 10,800

P = 17,500 W

Step 2: Calculate current

I = P / V = 17,500 / 220 = 79.5 A

Answer (b): The current drawn is 79.5 A.

Explanation:

The energy required to heat water depends on its mass, specific heat capacity, and temperature change. Because the system is not perfectly efficient, more electrical energy is needed than the actual heat gained by the water. The total electrical energy determines the cost, while the power and voltage determine the current used by the heater.

Example 19.18

An electric power station transmits 6.00 × 10⁷ W of electrical power through transmission lines at a voltage of 15.0 kV. The transmission distance is 1.20 km. The system is designed so that only 0.0200% of the power is lost as heat in the wires.

(a) What current is required to transmit this power?

(b) Determine the resistance of the transmission line that results in this power loss.

(c) Calculate the diameter of a copper wire of length 1.20 km that has this resistance. (Resistivity of copper ρ = 1.7 × 10^-8 Ω·m)

(d) What is unrealistic about the result?

(e) Which assumptions or conditions are not practical?

Jawab

Given:

• Power, $P = 6.00 \times 10^7 \ W$
• Voltage, $V = 15.0 \ kV = 1.5 \times 10^4 \ V$
• Length, $L = 1.20 \ km = 1200 \ m$
• Power loss fraction = $0.0200% = 2.0 \times 10^{-4}$
• Resistivity of copper, $\rho = 1.7 \times 10^{-8} \ \Omega m$

Concept:

Electrical power is given by:

$P = VI$

Power loss due to resistance:

$P_{loss} = I^2R$

Resistance of a wire:

$R = \rho \frac{L}{A}$

Area of cross-section:

$A = \pi r^2$

(a) Current required

$I = P/V$

$I = (6.00 \times 10^7)/(1.5 \times 10^4)$

I = 4.00 × 10³ A

Answer (a): The current is 4.00 × 10³ A.

(b) Resistance of the transmission line

Step 1: Calculate power loss

$P_{loss} = (2.0 \times 10^{-4}) \times (6.00 \times 10^7)$

$P_{loss} = 1.20 \times 10^4 \ W$

Step 2: Use $P_{loss}= I^2R$

$R = P_{loss}/I^2$

$R = (1.20 \times 10^4)/(4.00 \times 10^3)^2$

R = 7.50 × 10^-4 Ω

Answer (b): The resistance is 7.50 × 10^-4 Ω.

(c) Diameter of the copper wire

Step 1: Find cross-sectional area

$A = \rho L / R$

$A = (1.7 \times 10^{-8} \times 1200)/(7.50 \times 10^{-4})$

$A = 2.72 \times 10^{-2} \ m^2$

Step 2: Find radius

$r = \sqrt{\frac{A}{π}}$

$r = \sqrt{\frac{2.72}{3.14}}$

$r ≈ 0.093 \ m$

Step 3: Diameter

d = 2r = 0.186 m ≈ 18.6 cm

Answer (c): The diameter is approximately 18.6 cm.

(d) What is unrealistic?

The calculated wire diameter (about 18.6 cm) is extremely large and impractical for real transmission lines.

Answer (d): The wire size is unrealistically ضخ and not feasible in real-world applications.

(e) Unreasonable assumptions

• The allowed power loss (0.0200%) is too small, requiring an extremely low resistance.
• This leads to an impractically ضخ wire diameter.
• In real systems, higher voltages and acceptable losses are used to balance cost and efficiency.

Answer (e): The assumption of extremely low power loss is unrealistic and leads to impractical design requirements.

Explanation:

This problem demonstrates the importance of optimizing voltage and acceptable power loss in electrical transmission. Very low resistance requires very thick wires, which are expensive and impractical. In reality, engineers balance efficiency, cost, and feasibility by using high voltages and moderate losses.

Applications of Electrical Energy and Power Dissipation

Electrical energy and power dissipation are widely applied in daily life and modern technology. When electric current flows through a conductor, electrical energy is converted into other forms such as heat, light, or mechanical energy. Below are some important applications:

1. Electric Heaters

Electric heaters use the principle of power dissipation where electrical energy is converted into heat energy. Devices such as water heaters, electric stoves, and irons rely on the formula $P = I^2R$ to generate heat efficiently.

2. Incandescent Light Bulbs

In incandescent bulbs, electrical energy is converted into light and heat. The filament heats up due to resistance when current flows, producing visible light.

3. Electrical Appliances

Common household appliances such as kettles, toasters, and hair dryers use electrical power to perform useful work. The rate of energy consumption depends on the power rating of the device.

4. Power Transmission Systems

In electrical power transmission, energy loss occurs due to resistance in wires. This is known as power dissipation (P = I²R). Engineers minimize these losses by using high voltage and low current systems.

5. Fuse and Safety Devices

Fuses and circuit breakers operate based on heating effects of current. When excessive current flows, power dissipation increases and melts the fuse wire, protecting electrical circuits.

6. Industrial Heating

Industries use electrical energy for heating processes such as metal melting, welding, and drying. These applications rely on controlled power dissipation.

7. Electronic Devices

Electronic components like resistors dissipate power in the form of heat. Proper heat management is important to prevent damage and ensure efficiency.

Conclusion

Electrical energy and power dissipation play a crucial role in both domestic and industrial applications. Understanding these concepts helps improve energy efficiency and safety in electrical systems.

Conclusion

Electrical energy and power dissipation are fundamental concepts in physics that explain how electrical energy is converted into other forms such as heat, light, and mechanical energy. When an electric current flows through a conductor, part of the energy is dissipated due to resistance, mainly in the form of heat.

The relationship between voltage, current, and resistance is described by key formulas such as $P = VI$, $P = I^2R$, and $P = V^2/R$. These equations are essential for analyzing electrical circuits and determining energy efficiency.

Understanding electrical energy and power dissipation is important in designing safe and efficient electrical systems. It helps reduce energy losses, improve performance, and ensure proper functioning of electrical devices in both domestic and industrial applications.

Frequently Asked Questions (FAQ) about Electrical Energy and Power Dissipation

1. What is electrical energy?

Electrical energy is the energy transferred by electric charges when current flows through a conductor. It can be converted into other forms such as heat, light, or mechanical energy.

2. What is power dissipation?

Power dissipation is the process by which electrical energy is converted into heat in a conductor due to resistance when an electric current flows.

3. What is the formula for electrical power?

The main formulas for electrical power are:

• $P = IV$
• $P = I^2R$
• $P = \frac{V^2}{R}$

4. Why does power dissipation produce heat?

When electrons move through a conductor, they collide with atoms in the lattice. These collisions transfer energy to the atoms, causing them to vibrate and produce heat.

5. How can power loss be reduced in electrical systems?

Power loss can be reduced by using high voltage and low current transmission, as well as using conductors with low resistance.

6. What are examples of electrical energy applications?

Examples include electric heaters, light bulbs, electric stoves, and electronic devices, all of which convert electrical energy into useful forms.

7. What is the SI unit of electrical power?

The SI unit of electrical power is the watt (W), where 1 watt equals 1 joule per second.

Learn more: Explore related topics such as Ohm’s Law, electric current, and resistance to deepen your understanding.

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