Electricity: Joule's Heating Effect, Fuse & Power Ratings Class 10

When electric current surges through the thin filament of an incandescent light bulb, electrical potential energy is transformed into radiant thermal heat, causing the tungsten coil to glow white-hot at over $2500^\circ\text{C}$. In an electric room heater, nichrome coils convert amperes into room-warming warmth; in an electric safety fuse, excessive current melts a thin wire to prevent an electrical fire.

In CBSE Class 10 Science, Chapter 11 (Electricity), Joule's Law of Heating, the three mathematical equations of Electric Power ($P = VI = I^2R = V^2/R$), and Appliance Rating Numericals carry approximately $5$ to $7$ marks.

In this master guide, we derive Joule's law from first principles and conquer high-scoring board exam numericals.

What You Will Learn

• Derivation of Joule's Law of Heating from work and potential difference: $H = I^2Rt$
• The three mathematical forms of Electric Power: When to use $P = I^2R$ vs. $P = V^2/R$
• Solving Appliance Rating Problems (e.g., $220\text{ V}, 100\text{ W}$ bulb operated at $110\text{ V}$)
• Selecting the correct Fuse Wire Rating for domestic appliances
• Why tungsten is used exclusively for incandescent lamp filaments
• The physics of series vs. parallel heating power

1. Derivation of Joule's Law of Heating

Consider a current $I$ flowing through a resistor of resistance $R$ connected across a potential difference $V$ for time $t$:

Step-by-Step Derivation:

1. By definition of electric potential difference ($V = W / Q$): $\mathbf{W = V \times Q}$
2. By definition of electric current ($I = Q / t \implies Q = I \times t$): $W = V \times (I \times t) = V I t$
3. By Ohm's Law, substitute $V = I \times R$: $W = (I \times R) \times I \times t = \mathbf{I^2 R t}$
4. Assuming all electrical work done is converted into heat energy ($H = W$): $\mathbf{H = I^2 R t}$

Three Conditions of Joule's Law:

The heat produced in a resistor is:

1. Directly proportional to the square of current ($H \propto I^2$).
2. Directly proportional to the resistance ($H \propto R$).
3. Directly proportional to the time ($H \propto t$).

2. The Three Power Equations: Series vs. Parallel

                        Electric Power (P = Work / Time)
                                        |
       +--------------------------------+--------------------------------+
       |                                |                                |
    P = V × I                        P = I² × R                       P = V² / R
General Definition              USE IN SERIES CIRCUITS!          USE IN PARALLEL CIRCUITS!
(Current and Voltage given)     (Current I is CONSTANT)          (Voltage V is CONSTANT)
                                High Resistance = MORE Power!    Low Resistance = MORE Power!

Important: <u>In household circuits, appliances are connected in PARALLEL across a constant 220 V supply. Therefore, apply $P = V^2/R$! A 100 W bulb has LOWER resistance than a 40 W bulb ($R = V^2/P$), drawing more current and glowing brighter!</u>

3. High-Yield Solved Board Examination Problems

Problem 1: Appliance Voltage Shift Numerical (CBSE Classic)

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

Solution:

1. Find the Resistance ($R$) of the Bulb: The resistance of an appliance is an intrinsic physical property that remains constant regardless of applied voltage! $P = \frac{V^2}{R} \implies R = \frac{V^2}{P}$ Substitute rating values: $R = \frac{(220)^2}{100} = \frac{48400}{100} = \mathbf{484\ \Omega}$
2. Calculate Power Consumed at $110\text{ V}$ ($P'$): $P' = \frac{(V')^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = \mathbf{25\text{ Watts}}$ (Shortcut: Since $V$ is halved ($110 = 220/2$), power drops by $(1/2)^2 = 1/4$ of $100\text{ W} = 25\text{ W}$!)
3. Therefore, <u>the power consumed when operated on $110 ext{ V}$ is $25 ext{ Watts}$</u>.

Problem 2: Selecting an Appropriate Electric Fuse

Problem: An electric iron consumes energy at a rate of $840\text{ W}$ when heating is at the maximum rate and $360\text{ W}$ when heating is at the minimum. The voltage is $220\text{ V}$. What is the current in each case, and what fuse rating should be selected?

Solution:

1. Case 1: Maximum Rate ($P = 840\text{ W}$): $I_1 = \frac{P}{V} = \frac{840\text{ W}}{220\text{ V}} = \mathbf{3.82\text{ Amperes}}$
2. Case 2: Minimum Rate ($P = 360\text{ W}$): $I_2 = \frac{P}{V} = \frac{360\text{ W}}{220\text{ V}} = \mathbf{1.64\text{ Amperes}}$
3. Select Fuse Rating:
   • The maximum normal operating current is $3.82\text{ A}$.
   • Standard commercial fuse ratings are $1\text{ A}, 2\text{ A}, 3\text{ A}, 5\text{ A}, 10\text{ A}$.
   • A $3\text{ A}$ fuse would blow under normal maximum operation.
   • Therefore, a $5\text{ A}$ fuse must be selected to protect the circuit safely!

4. Why Tungsten is Used for Bulb Filaments

1. Very High Melting Point: Tungsten has an extraordinarily high melting point of $3380^\circ\text{C}$, enabling it to glow white-hot without melting.
2. High Electrical Resistivity: Generates intense Joule heating ($H = I^2Rt$).
3. Inert Gas Protection: Bulbs are filled with chemically unreactive gases like Argon ($Ar$) and Nitrogen ($N_2$) to prevent the hot tungsten filament from oxidizing and burning.

5. Summary and Examination Tips

Exam Tip: In questions asking "Why does the cord of an electric heater not glow while the heating element does?": State that the connecting cord (copper) has very low resistance ($H \propto R$), producing negligible heat, while the heating element (nichrome) has very high resistance, producing intense glowing heat!

Common Mistake: Thinking power halves when voltage halves. Because $P = V^2/R$, halving voltage reduces power to one-fourth ($25\%$)!