In internal wiring estimation, if the connected load is 2 kW a... | Technical Ability MCQ | Examoogle

Technical AbilityElectrical Installations and Estimation

PrevNext

In internal wiring estimation, if the connected load is 2 kW and the supply voltage is 240 V, then the maximum load current will be ________.

4.33 A

1.14 A

80 A

8.33 A

Correct Answer

⚙️ TE • Technical Ohm's Law ApplicationTechnical AbilityElectrical Installations and Estimation

Option D

8.33 A

Quick Summary:

In an electrical circuit, the load current $I$ is calculated by dividing the total power $P$ by the voltage $V$ . Given a load of $2\text{ kW}$ ( $2000\text{ W}$ ) and a supply voltage of $240\text{ V}$ , the current is determined using the relation $I = \frac{P}{V}$ .

⚙️TETechnical SolutionOhm's Law Application

📜 Standard Reference & Code Clause

IS 732:1989 (Code of Practice for Electrical Wiring Installations)

💡 Explanation

🔢 Key Formulas

$I = \frac{P}{V}$ — formula to calculate current in Amperes where $P$ is in Watts and $V$ in Volts

$P = V \times I$ — relationship for DC or unity power factor AC circuits

⚙️ Working Principle

Current is the flow of electrons through a conductor driven by the potential difference $V$ . In resistive loads (like lighting and heating circuits typically assumed in estimation), the power $P$ consumed is directly proportional to the voltage applied and the resulting current $I$ . As $V$ increases, $I$ decreases for a constant $P$ , governed by $I = \frac{P}{V}$ .

📌 Key Points

▸
Always convert kiloWatts (kW) to Watts (W) before division: $2\text{ kW} = 2000\text{ W}$ .
▸
Current is measured in Amperes ( $A$ ).
▸
In domestic estimation, the supply voltage is standard at $240\text{ V}$ AC for single-phase systems.

✅ Advantages

▸
Simple direct calculation.
▸
Essential for sizing miniature circuit breakers (MCB).

❌ Disadvantages / Limitations

▸
Assumes purely resistive load (ignoring power factor).
▸
Does not account for starting surge currents of inductive loads.

🛠️ Applications / Uses

▸
Residential circuit design.
▸
Selection of cable cross-sectional area.
▸
Determination of protective device ratings.

📄 Additional Information

▸
The calculation $2000 \div 240 = 8.333...$ leads to $8.33\text{ A}$ .
▸
Option A is the result of using a $120\text{ V}$ assumption (a common error).
▸
Option B is the result of inverse calculation ( $240 \div 2000 = 0.12$ ).
▸
Option C is significantly higher, likely a calculation error involving magnitude.

📊 Diagram / Illustration

✅

D is correct — The current is calculated as $I = \frac{2000}{240} \approx 8.33\text{ A}$ .

Core Concepts Used

Click any tag to open in AI Tutor

Power consumption Ohm's law Electrical estimation

💡 EXAM TIP

Always ensure units match before calculation. If the power were given in kVA, you would need to multiply by the power factor ( $\cos \phi$ ) to find the real power in Watts.

Was this solution helpful?

⚡ Practice on Android App · 4.8 ★

Solve 10,000+ More Questions with AI Tutor

Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.

Free Download • 100K+ Aspirants

Discussion (0)

Loading discussion...

PrevNext

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)