Final answer:
The peak current through an inductor is 15.5 mA when connected to an AC source with a peak voltage of 0.90 V is approximately 0.909 A. The correct answer is d) None of these
Step-by-step explanation:
The peak current through an inductor can be found using the formula:
Ipeak = Vpeak / XL
Where
Ipeak is the peak current
Vpeak is the peak voltage
XL is the reactance of the inductor.
Since the reactance of an inductor is given by XL = 2πfL, where f is the frequency and L is the inductance
We can rearrange the formula to solve for the peak current:
Ipeak = Vpeak / (2πfL)
Substituting the given values: Vpeak = 0.90 V, f = 60 Hz, and L = 7.0 mH (convert to H by dividing by 1000), we can calculate:
Ipeak = 0.90 V / (2π * 60 Hz * 7.0 * 10-3 H)
Ipeak ≈ 0.909 A
Therefore, the correct answer is d) None of these options as none of them match the calculated value.