Q159 / 180 · COMEDK 2023 (Evening Shift) · Physics
What should be the inductance of an inductor connected to $200 \mathrm{~V}, 50 \mathrm{~Hz}$ source so that the maximum current of $\sqrt{2}$ A flows through it?
- A$\dfrac{\pi}{2} H$
- B$2 \pi H$
- C$\dfrac{\sqrt{2}}{\pi} H$
- Correct answerD$\dfrac{2}{\pi} \mathrm{H}$
Solution
The root mean square voltage is given as $V_{rms} = 200 \text{ V}$. The maximum current is given as $I_{max} = \sqrt{2} \text{ A}$.
The root mean square current is $I_{rms} = \dfrac{I_{max}}{\sqrt{2}} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1 \text{ A}$.
The inductive reactance $X_L$ is given by $X_L = \dfrac{V_{rms}}{I_{rms}} = \dfrac{200}{1} = 200 \text{ } \Omega$.
The formula for inductive reactance is $X_L = \omega L = 2 \pi f L$, where $f = 50 \text{ Hz}$.
Substituting the values: $200 = 2 \times \pi \times 50 \times L$.
$200 = 100 \pi L$.
$L = \dfrac{200}{100 \pi} = \dfrac{2}{\pi} \text{ H}$.
Answer: $\dfrac{2}{\pi} \mathrm{H}$
The root mean square current is $I_{rms} = \dfrac{I_{max}}{\sqrt{2}} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1 \text{ A}$.
The inductive reactance $X_L$ is given by $X_L = \dfrac{V_{rms}}{I_{rms}} = \dfrac{200}{1} = 200 \text{ } \Omega$.
The formula for inductive reactance is $X_L = \omega L = 2 \pi f L$, where $f = 50 \text{ Hz}$.
Substituting the values: $200 = 2 \times \pi \times 50 \times L$.
$200 = 100 \pi L$.
$L = \dfrac{200}{100 \pi} = \dfrac{2}{\pi} \text{ H}$.
Answer: $\dfrac{2}{\pi} \mathrm{H}$
Hence, the correct answer is option: 'D'.