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Technical AbilityBasic Electrical Engineering

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The value of inductance needed to store 4kWh of energy in a coil carrying a 2000A current is:

72 H

720 H

7.2 $\times 10$ ⁶ H

7.2 H

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityBasic Electrical Engineering

Option D

7.2 H

Quick Summary:

The energy stored in an inductor is proportional to the square of the current flowing through it. To find the inductance, we equate the given energy in joules to the formula $E = \frac{1}{2} L I^2$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

The energy stored in an inductor is proportional to the square of the current flowing through it. To find the inductance, we equate the given energy in joules to the formula $E = \frac{1}{2} L I^2$ .

🔢 Key Formulas

$E = \frac{1}{2} L I^2$ — Formula for energy stored in an inductor where $E$ is energy in Joules, $L$ is inductance in Henry, and $I$ is current in Amperes.

⚙️ Working Principle

When current flows through a coil, it creates a magnetic field. The energy is stored within this magnetic field. As the current increases, the magnetic flux increases, leading to an accumulation of potential energy $E = \int v i dt$ , which results in the expression $\frac{1}{2} L I^2$ .

📌 Key Points

▸
Energy must be converted to SI units: $4 \text{ kWh} = 4 \times 10^3 \times 3600 \text{ J} = 14.4 \times 10^6 \text{ J}$ .
▸
Given current $I = 2000 \text{ A}$ .
▸
Rearranging the formula gives $L = \frac{2E}{I^2}$ .

✅ Advantages

▸
Energy density management in power systems
▸
Fundamental for inductive components in filter circuits

❌ Disadvantages / Limitations

▸
High currents lead to significant heating due to coil resistance
▸
Inductors cannot change current instantaneously

🛠️ Applications / Uses

▸
Energy storage systems in SMES (Superconducting Magnetic Energy Storage)
▸
Inductive filtering in DC-DC converters

📄 Additional Information

▸
Calculation: $L = \frac{2 \times 14.4 \times 10^6}{(2000)^2} = \frac{28.8 \times 10^6}{4 \times 10^6} = 7.2 \text{ H}$ .
▸
The energy must be strictly converted from kWh (kilowatt-hour) to Joules (watt-seconds) before substitution to ensure unit consistency.

📊 Diagram / Illustration

✅

D is correct — The inductance required to store $14.4 \times 10^6 \text{ J}$ of energy at $2000 \text{ A}$ is calculated to be $7.2 \text{ H}$ .

Core Concepts Used

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Magnetic Energy Storage Electromagnetic Induction Unit Conversion (kWh to Joules)

💡 EXAM TIP

Always verify unit consistency in electrical engineering problems; energy storage formulas typically use Joules, so convert hours to seconds and kilowatts to watts immediately.

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