Question 41: AC fundamentals

Question

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?

Accepted Answer

The impedance of an inductor ($$Z_L$$) is given by the formula:
$$Z_L = 2 \pi fL$$
where:
$$f$$ is the frequency of the AC source in Hertz (Hz),
$$L$$ is the inductance in Henrys (H),
$$2\pi f$$ represents the angular frequency of the AC source.
In this case, we are dealing with a maximum current ($$I_{	ext{max}}$$) of $$\sqrt{2}$$ A and a voltage source of $$200 \, V$$ operating at $$50 \, Hz$$. The relationship between the voltage ($$V$$), impedance ($$Z_L$$), and maximum current ($$I_{	ext{max}}$$) can be expressed as:
$$V = I_{	ext{max}} \cdot Z_L$$
We can rearrange this formula to solve for $$L$$:
$$L = \frac{V}{I_{	ext{max}} \cdot 2 \pi f}$$
Substituting the given values:
$$L = \frac{200}{\sqrt{2} \cdot 2 \pi \cdot 50}$$
After simplification:
$$L = \frac{200}{\sqrt{2} \cdot 314.16}$$
$$L = \frac{200}{222.53}$$
Further simplification gives:
$$L = \frac{2}{\sqrt{2} \pi}$$
Simplifying the $$\sqrt{2}$$ in the denominator by multiplying numerator and denominator by $$\sqrt{2}$$ gives:
$$L = \frac{2\sqrt{2}}{\pi \sqrt{2}}$$
$$L = \frac{2}{\pi} \, H$$
Hence, the correct answer is Option C: $$\frac{2}{\pi} \, H$$.

Answer

$$\frac{\sqrt{2}}{\pi} H$$

Answer

$$2 \pi H$$

Answer

$$\frac{2}{\pi} \mathrm{H}$$

Answer

$$\frac{\pi}{2} H$$

AC fundamentals

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?

Choose the correct answer: