Question

If the current passing through an inductor is increased by 20%, then the percentage increase in the energy stored in the inductor is

• A. 44
• B. 36
• C. 20
• D. 10

Hint:

For small percentage changes ($<5%$), you can use calculus: $\frac{\Delta U}{U} \approx 2 \frac{\Delta I}{I}$. Here, the change is 20%, which is large, so use the exact method.

Answer 1

The Correct Option is A

Step 1: Formula for Energy Stored:
The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$. 

Step 2: Calculate New Energy:
Let initial current be $I$. New current increases by $20\%$: $$ I' = I + 0.20I = 1.2I $$ New energy $U'$: $$ U' = \frac{1}{2} L (I')^2 = \frac{1}{2} L (1.2I)^2 $$ $$ U' = \frac{1}{2} L (1.44 I^2) $$ $$ U' = 1.44 \left( \frac{1}{2} L I^2 \right) = 1.44U $$ 
Step 3: Calculate Percentage Increase:
$$ \% \text{ Increase} = \frac{U' - U}{U} \times 100 $$ $$ \% \text{ Increase} = \frac{1.44U - U}{U} \times 100 $$ $$ \% \text{ Increase} = 0.44 \times 100 = 44\% $$ 
Step 4: Final Answer:
The percentage increase in energy is 44\%.