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Nady [450]
3 years ago
5

An inductor connects to an AC circuit. If the frequency of the AC oscillations decreases, does the inductance increase, decrease

, or remain constant?
Physics
1 answer:
ASHA 777 [7]3 years ago
7 0

Answer:

The inductance remains constant

Explanation:

The inductance value remains constant because it is a intrinsic property of an inductor. Its value depends on physical characteristics and dimensions like length, number of turns, etc.

What changes is the impedance of the inductance. Now, the impedance does changes with the frequency of the AC signal. The relations is as follows:

Z = jωL, where Z is the impedance, j is the imaginary unit, ω is the angular frequency, and L is the inductance.

In terms of frequency ω=2π·f, where f is the frequency of the signal.

So, the impedance in terms of f is:

Z = j·2·π·f·L

If f decreases, the value of the Z decreases too, because if f has a lower value and the other terms remain constant, the value of Z will be lower.

So, the inductance remains constant.

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Mr. Adams asked his students to write the chemical symbol for the element zinc. He wrote the students’ answers on the whiteboar
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2 years ago
The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w
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Explanation:

The given data is as follows.

         Resistance (R) = 1200 ohm,     Area (A) = 20 \times 10^{-6} m (as 1 \mu m = 10^{-6} m)

            Diameter (d) = 2.3 mm = 2.3 \times 10^{-3} m

First, we will calculate the length as follows.

            R = \rho \frac{L}{A}

Here,  \rho = resistivity of aluminium = 2.65 \times 10^{-8}

Putting the given values above and we will calculate the value of length as follows.

               R = \rho \frac{L}{A}

             1200 = 2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}

               L = 9.056 \times 10^{5}

As the circumference of circular wire = 2 \pi r

or,                                                          = 2 \times \pi \times \frac{d}{2}  

                                                              = \pi \times d

And, number of turns will be calculated as follows.

             No. of turns × Circumference = Length of wire

              No. of turns × 3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}

                               = 1.25 \times 10^{8}

Thus, we can conclude that 1.25 \times 10^{8}  turns of wire are needed.

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3 years ago
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