The peak current to and from an inductor is i when the emf source has a peak voltage V and frequency f. What is the peak current to and from the inductor if the frequency and peak voltage are doubled to 2f and 2V$2\mathrm{V}$, respectively?

• A. i/4
• B. i/2
• C. 2i
• D. 4i
• E. i

Answer: (E)

Verified step by step guidance

1. Understand that the relationship between voltage, current, and inductance in an AC circuit is given by the formula: \( V = I \cdot X_L \), where \( X_L = 2\pi f L \) is the inductive reactance.
2. Recognize that when the frequency \( f \) and the peak voltage \( V \) are doubled, the new frequency becomes \( 2f \) and the new peak voltage becomes \( 2V \).
3. Calculate the new inductive reactance using the formula \( X_L = 2\pi (2f) L = 4\pi f L \). This shows that the inductive reactance is doubled.
4. Substitute the new values into the voltage-current relationship: \( 2V = I' \cdot 4\pi f L \).
5. Solve for the new peak current \( I' \) by rearranging the equation: \( I' = \frac{2V}{4\pi f L} = \frac{V}{2\pi f L} = \frac{i}{2} \). Thus, the peak current is halved, confirming the correct answer is \( i \).