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Amanda [17]
2 years ago
11

An important concept to aid understanding of electromagnetics is electrical length. Electrical _________ is a unitless measure t

hat refers to the length of a wire or device at a certain ___________ . It is defined as the ration of the physical ___________ of the device to the_____________ of the signal frequency.
Engineering
1 answer:
kondor19780726 [428]2 years ago
3 0

Answer: Electrical length is a unitless measure that refers to the length of a wire or device at a certain frquency. It is defined as the ratio of the physical length of the device to the wavelength of the signal frequency.

Explanation:

In nature, the maximum speed achievable for any interaction or perturbation along a medium (like a wire) can't be higher than the speed of light, otherwise it would be violating Einstein's Relativity Theory.

So, in a given circuit, the voltage and current aren't set up instantaneously, and a finite time exists since a battery is connected to a circuit till a stable current  traverses a load resistor connected to it.

Now, the electrical length, defines how important in is this delay, in the calculation of the voltage through the resistor, for instance.

We can write the electrical length (L.E.) as follows:

L.E. = L / (v/f) = L / λ

As we can see, the effect depends not only on the circuit dimensions, but the frequency of the signal as well.

If L.E. is much smaller than λ, this means that we can neglect the effects of the delay, and we can use the circuit theory, namely KVL, KCL and Ohm's law to analyze the circuit, assuming that the current establishes instantaneously.

Otherwise, the current and voltage concepts are not valid anymore, as we need to think in terms of propagation of electromagnetic waves, like in transmission lines and antennas.

Roughly, if L.E. ≤ 20 λ, we say that we are still in the realm of the circuit theory.

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6 0
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A diagonal aluminum alloy tension rod of diameter d and initial length l is used in a rectangular frame to prevent collapse. The
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Answer:

\Delta L = 0.1883 inch

Explanation:

Given data:

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\sigma_{allow} = 20  kpsi

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              = \frac{P}{A}\times \frac{L}{E}

              = \sigma_{allow} \times \frac{L}{E}

modulus of elasticity E for aluminium alloy is 10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi

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Answer:true

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