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hodyreva [135]
3 years ago
14

A certain metal has a resistivity of 1.68 × 10-8 Ω ∙ m. You have a long spool of wire made from this metal. If this wire has a d

iameter of 0.15 mm, how long should you cut a segment so its resistance will be 15 Ω?
Engineering
1 answer:
Vesna [10]3 years ago
4 0

Answer:

15.78 m

Explanation:

Length = resistance×area/resistivity

Resistance = 15 ohms

Area = πd^2/4 = 3.142×(0.15/1000)^2/4 = 1.767375×10^-8 m^2

Resistivity = 1.68 ohm-meter

Length = 15×1.767375×10^-8/1.68×10^-8 = 15.78 m

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Answer:

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To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin⁡(ωt) , is flowing through the indu
Sophie [7]

Answer:

V(t) = XLI₀sin(π/2 - ωt)

Explanation:

According to Maxwell's equation which is expressed as;

V(t) = dФ/dt ........(1)

Magnetic flux Ф can also be expressed as;

Ф = LI(t)

Where

L = inductance of the inductor

I = current in Ampere

We can therefore Express Maxwell equation as:

V(t) = dLI(t)/dt ....... (2)

Since the inductance is constant then voltage remains

V(t) = LdI(t)/dt

In an AC circuit, the current is time varying and it is given in the form of

I(t) = I₀sin(ωt)

Substitutes the current I(t) into equation (2)

Then the voltage across inductor will be expressed as

V(t) = Ld(I₀sin(ωt))/dt

V(t) = LI₀ωcos(ωt)

Where cos(ωt) = sin(π/2 - ωt)

Then

V(t) = ωLI₀sin(π/2 - ωt) .....(3)

Because the voltage and current are out of phase with the phase difference of π/2 or 90°

The inductive reactance XL = ωL

Substitute ωL for XL in equation (3)

Therefore, the voltage across inductor is can be expressed as;

V(t) = XLI₀sin(π/2 - ωt)

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