🔥 11.7 Heating Effect of Electric Current
We know that a battery or a cell is a source of electrical energy. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to flow the current through a resistor or a system of resistors connected to the battery.
Where does this energy go? A part of the source energy in maintaining the current may be consumed into useful work (like in rotating the blades of an electric fan). Rest of the source energy may be expended in heat to raise the temperature of gadget.
H = VIt
For a steady current I, the amount of heat H produced in time t is given by the above formula.
⚡ Joule's Law of Heating
Applying Ohm's law [V = IR], we get:
H = I²Rt
This is known as Joule's law of heating. The law implies that heat produced in a resistor is:
- (i) directly proportional to the square of current for a given resistance
- (ii) directly proportional to resistance for a given current
- (iii) directly proportional to the time for which the current flows through the resistor
💡 Practical Applications of Heating Effect
The generation of heat in a conductor is an inevitable consequence of electric current. In many cases, it is undesirable as it converts useful electrical energy into heat. However, heating effect of electric current has many useful applications.
Electric Bulb
The filament must retain as much of the heat generated as possible, so that it gets very hot and emits light. Tungsten (melting point 3380°C) is used for making bulb filaments. Bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of filament.
Electric Fuse
It protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point (aluminium, copper, iron, lead etc.). Fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc.
⚡ 11.8 Electric Power
The rate of doing work is power. This is also the rate of consumption of energy.
P = VI = I²R = V²/R
The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.
Units:
- 1 W = 1 volt × 1 ampere = 1 V A
- 1 kilowatt (kW) = 1000 watts
- 1 kilowatt hour (kWh) = 3.6 × 10⁶ joule (J)
Example 11.10 - Electric Iron Power Analysis
Problem: An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
Solution:
From P = V I, we know that I = P/V
(a) When heating is at the maximum rate:
I = 840 W / 220 V = 3.82 A
R = V/I = 220 V / 3.82 A = 57.60 Ω
(b) When heating is at the minimum rate:
I = 360 W / 220 V = 1.64 A
R = V/I = 220 V / 1.64 A = 134.15 Ω
Example 11.11 - Heat and Potential Difference
Problem: 100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
Solution:
H = 100 J, R = 4 Ω, t = 1 s, V = ?
From H = I²Rt: I = √(H/Rt) = √[100 J/(4 Ω × 1 s)] = 5 A
Thus V = IR = 5 A × 4 Ω = 20 V
Example 11.12 - Power of a Bulb
Problem: An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Solution:
P = VI
P = 220 V × 0.50 A
P = 110 J/s = 110 W
Example 11.13 - Electricity Bill Calculation
Problem: An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
Solution:
Total energy = 400 W × 8.0 hour/day × 30 days = 96000 W h
= 96 kW h
Cost = 96 kW h × Rs 3.00 per kW h = Rs 288.00
Practice Questions
- Why does the cord of an electric heater not glow while the heating element does?
- Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
- An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
- What determines the rate at which energy is delivered by a current?
- An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Question 1: Why does the cord of an electric heater not glow while the heating element does?
Let's understand this concept through visualization and analysis.
Question 2: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
We'll solve this step-by-step using the formula H = QV.
Question 3: An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
We'll apply Joule's law to find the heat produced.
Question 4: What determines the rate at which energy is delivered by a current?
Let's explore the concept of electric power and its determining factors.
Question 5: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
We'll calculate both power and energy consumption step by step.
| Quantity |
Symbol |
Unit |
Formula |
| Heat |
H |
Joule (J) |
H = I²Rt = VIt |
| Electric Power |
P |
Watt (W) |
P = VI = I²R = V²/R |
| Energy |
W |
Joule (J) |
W = P × t |
| Commercial Energy |
- |
kWh (unit) |
1 kWh = 3.6 × 10⁶ J |