Question

In: Physics

An inductor has a 43.6 Ω inductive reactance when connected across a 60.0 Hz power source....

An inductor has a 43.6 Ω inductive reactance when connected across a 60.0 Hz
power source. The inductor is removed from this first power source and then
connected to a 52.0 Hz power source that produces a 80.0 Vrms voltage.
a. (5 points) Draw a clear physics diagram of the problem.
b. (4 points) What is the inductance for this inductor?
c. (6 points) What is the maximum current in the inductor when connected to the
52.0 Hz source?
d. (5 points) Suppose a capacitor is now connected in series with the inductor from
part (c), with the same power source. Would the maximum current in the inductor
increase, decrease, or stay

Solutions

Expert Solution

Purely Inductive AC Circuit:

In a completely inductive AC circuit, the voltage and current have a 90-degree difference in their phases and the current is ahead of the voltage. In addition, the value of the reactance in the inductor is influenced by the value of the frequency of the circuit voltage source. Therefore, if we vary the frequency of the source, the value of the current varies and so does its reactance.

Given:-

Reactance(XL)=43.6 ohm

Frequency(f)=60 Hz

Voitage(v) = 80 Vrms

Frequncy (f)=52 Hz

a.

b. The inductance for this inductor is:

We have that the inductive reactance is XL =43.6 ohm , when f= 60.0Hz.

If f=60Hz the angular frequency will be:

Then, the inductive reactance is given by:

where L is the inductance in Henry.

Solving for L:

c. Now, when the frequency is f= 52 Hz

Thus, the inductive reactance is:

The maximum source voltage is determined by:

The maximum current in the circuit, when the AC source produce Vmax, is:

Therefore, the maximum current in the inductor, when connected to the
52.0 Hz source is Imax = 3.01 A.

d. When you close the switch and let current flow out of the capacitor, it can't flow right away because the rapidly changing current sets up an opposing voltage in the inductor or at resonance condition where XL=XC, the current increases slowly, reaching a maximum and when the capacitor is fully discharged, the current starts decreasing, which induces an opposite current in the inductor, thus, in result, the capacitor gets charged conversely.


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