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OlgaM077 [116]
3 years ago
12

What is a reasonable approximation for an inductor at DC steady state?

Engineering
1 answer:
abruzzese [7]3 years ago
7 0

Answer:

Short circuit

Explanation:

An inductor works on the principle of electromagnetic induction.

When a current passes through an inductor, a magnetic field is generated inside it; if the magnitude of this magnetic field changes, then according to Faraday-Newmann-Lenz law, an emf (electromotive force) is induced in the inductor itself, according to:

\epsilon=-\frac{d\Phi}{dt}

where \Phi is the magnetic flux linkage through the inductor. Therefore, as a result of this induced emf, a current is also induced in the inductor, and the direction of this current is such that it opposes to the change in magnetic flux linkage.

Therefore, the intensity of this current depends on the rate of change of magnetic flux linkage through the inductor.

If we have a DC (direct current), then the current in the inductor is steady; as a result, its magnetic field is also steady, and therefore, there is no change in magnetic flux, so no induced current. Therefore, the inductor will oppose no resistance to the flow of original current, and so it can be approximated to a short circuit (zero resistance).

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Answer:

percentage change in volume  = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × 10^{11}  N/m²

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solution

we will apply here bulk modulus formula that is

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put here value and we get

3.5 × 10^{11} = \frac{10^7}{bulk\ strain}  

solve it we get

bulk strain = 2.8571 × 10^{-5}

and

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so that percentage change in volume is = 2.8571 × 10^{-5}  × 100

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