A current of 250 mA flows through a resistor of 10 Ω. Calculat... | General Science MCQ | Examoogle

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A current of 250 mA flows through a resistor of 10 Ω. Calculate the voltage across the resistor.

2.5 V

25 V

0.25 V

250 V

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral ScienceGeneral Science

Option A

5 V

Quick Summary:

To calculate the voltage, Ohm's Law is applied by multiplying the current by the resistance. Given $I = 250 \text{ mA} = 0.25 \text{ A}$ and $R = 10 \text{ } \Omega$ , the voltage $V = I \times R = 0.25 \text{ A} \times 10 \text{ } \Omega = 2.5 \text{ V}$ .

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🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of voltage like water pressure in a pipe, current like the volume of water flowing, and resistance like the narrowness of the pipe; to get the same flow through a narrower pipe, you need more pressure.

🧠 Memory Mnemonic / Shortcut Aid

Remember 'VIR': V on top of I and R in a triangle ( $V = I \times R$ , $I = \frac{V}{R}$ , $R = \frac{V}{I}$ ).

🔢 Key Formulas

$V = I \times R$ — Ohm's Law relationship between Voltage ( $V$ ), Current ( $I$ ), and Resistance ( $R$ )

$I_{converted} = \frac{I_{mA}}{1000}$ — Conversion from milliamperes to amperes

⚙️ Working Principle

Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. The proportionality constant is the electrical resistance ( $R$ ). Mathematically, this is expressed as $V = I \times R$ , where $V$ is potential difference in volts, $I$ is current in amperes, and $R$ is resistance in ohms.

📌 Key Points

▸
Always convert milliamperes (mA) to amperes (A) before performing calculations in standard units.
▸
The SI unit of potential difference is the Volt (V), and resistance is measured in Ohms ( $\Omega$ ).
▸
Ohm's Law is only applicable to ohmic conductors under constant physical conditions.

✅ Advantages

▸
Simple linear relationship for analysis of electrical circuits.

❌ Disadvantages / Limitations

▸
Does not apply to non-ohmic devices like diodes or transistors where $V$ is not proportional to $I$ .

🛠️ Applications / Uses

▸
Circuit design and troubleshooting in electronics.
▸
Determining power consumption in household appliances.

📄 Additional Information

▸
The prefix 'milli' denotes a factor of $10^{-3}$ , so $250 \text{ mA} = 250 \times 10^{-3} \text{ A} = 0.25 \text{ A}$ .
▸
Option B (25 V) results from failing to convert mA to A. Option C (0.25 V) arises from misapplying the formula as $I \div R$ .

📊 Diagram / Illustration

✅

A is correct — The voltage across the resistor is calculated using Ohm's Law as $V = I \times R$ , which equals $0.25 \text{ A} \times 10 \text{ } \Omega = 2.5 \text{ V}$ .

Core Concepts Used

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Ohm's Law Electrical Potential Difference Unit Conversion

💡 EXAM TIP

In competitive exams, always check unit consistency first; ignoring the 'm' in 'mA' is a classic trap leading to incorrect orders of magnitude.

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