Question

A 15.0 mh inductor is connected across an ac generator that produces a peak voltage of 9.00 v.

what is the peak current through the inductor if the emf frequency is 100 hz ?

Answer 1

The peak current through the inductor is 0.957 A.

Step-by-step explanation:

To calculate the peak current through the inductor, we can use the formula:

Imax = Vm / XL

Vm represents the peak voltage and XL represents the inductive reactance. In this case, Vm is given as 9.00 V and the inductor has an inductance of 15.0 mH (or 0.015 H). The formula for inductive reactance is:

XL = 2πfL

Substituting the given values into the formula, we have:

XL = 2π(100 Hz)(0.015 H) = 9.42 Ω

Now we can calculate the peak current:

Imax = (9.00 V) / (9.42 Ω) = 0.957 A

Answer 2

The peak current through the inductor is approximately 0.95 A.

How to solve

Peak current through the inductor can be calculated using the formula:

`I_peak = V_peak / (2 * \pi * f * L)`

where:

V_peak is the peak voltage (9.00 V)

f is the frequency (100 Hz)

L is the inductance (15.0 mH)

Plugging in the values:

`I_(peak) = 9.00 V / (2 * \pi * 100 Hz * 15.0 mH) = 0.95 A`

Therefore, the peak current through the inductor is approximately 0.95 A.