A potential difference of 250 volts is applied across a resist... | General Science MCQ | Examoogle

General SciencePhysics: Electricity

A potential difference of 250 volts is applied across a resistance of 500 ohms in an electric iron. Calculate heat energy produced in joules in 10 seconds.

125

500

250

1250

None of the above

Correct Answer

⚙️ TE • Technical Concept & PrincipleGeneral SciencePhysics: Electricity

Option D

1250

Quick Summary:

The heat energy produced in an electrical conductor is governed by Joule's Law of Heating. Given the potential difference $V = 250\text{ V}$ , resistance $R = 500\text{ }\Omega$ , and time $t = 10\text{ s}$ , the heat energy $H$ is calculated as $H = \frac{V^2}{R} \times t$ .

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of electrical resistance like a narrow corridor; pushing a crowd of people (electrons) through it causes friction and heat, similar to how an electric iron heats up by resisting the flow of current.

🧠 Memory Mnemonic / Shortcut Aid

VIT (V square over R, times T) is a way to remember $H = \frac{V^2}{R}t$ .

🔢 Key Formulas

$H = \frac{V^2}{R} \times t$ — Joule's law of heating in terms of voltage, resistance, and time

$P = \frac{V^2}{R}$ — Formula for electrical power

⚙️ Working Principle

When a current flows through a conductor with resistance, electrical energy is converted into thermal energy due to the collisions between moving electrons and the lattice ions of the conductor. The rate of energy dissipation is defined by the power formula $P = \frac{V^2}{R}$ . Total heat energy is obtained by multiplying this power by the duration of flow.

📌 Key Points

▸
Joule's Law states that heat produced is directly proportional to the square of the current, resistance, and time.
▸
The SI unit of heat energy is Joule (J).
▸
Voltage and resistance are inversely related in the power dissipation formula.

✅ Advantages

▸
Allows for controlled heating in appliances like irons, heaters, and toasters.

❌ Disadvantages / Limitations

▸
Energy loss as heat in transmission lines is a major concern in power distribution.

🛠️ Applications / Uses

▸
Electric iron for pressing clothes
▸
Electric fuse for circuit protection

📄 Additional Information

▸
Calculation: $H = \frac{250^2}{500} \times 10 = \frac{62500}{500} \times 10 = 125 \times 10 = 1250\text{ J}$ .
▸
Option A is the result of using incorrect values; Option B is $V \times t / R$ incorrectly applied; Option C is $V^2 / R$ without time.

📊 Diagram / Illustration

✅

D is correct — The calculation $H = \frac{250^2}{500} \times 10$ yields exactly 1250 Joules of heat energy.

Core Concepts Used

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Joule's Law of Heating Ohm's Law Electrical Power

💡 EXAM TIP

Always verify units before calculation; ensure voltage is in Volts, resistance in Ohms, and time in Seconds to get energy in Joules.

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