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irina [24]
3 years ago
10

In a parallel circuit, ET = 120 V, R = 470 Ω, and XL = 500 Ω. What is the reactive power?

Mathematics
1 answer:
ddd [48]3 years ago
8 0

Answer:

Reactive power is 28.8VARs

Step-by-step explanation:

We are given that in a parallel circuit, E_{T}=120V,R=470Ω and X_{L}=500Ω

And we area asked to find reactive power.

Reactive power is power dissipated due to reactive elements in the circuit and it occurs when voltage and current are not in phase.

Reactive power = \frac{E_{T}^{2}}{X_{L}}=\frac{120^{2}}{500}=28.8VARs

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<h3>Procedure - Determination of an appropriate function based on given information</h3>

In this question we must find an appropriate model for a <em>periodic</em> function based on the information from statement. <em>Sinusoidal</em> functions are the most typical functions which intersects a midline (x_{mid}) and has both a maximum (x_{max}) and a minimum (x_{min}).

Sinusoidal functions have in most cases the following form:

x(t) = x_{mid} + \left(\frac{x_{max}-x_{min}}{2} \right)\cdot \sin (\omega \cdot t + \phi) (1)

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If we know that x_{min} = -14.5, x_{mid} = -8, x_{max} = -1.5, (t, x) = (-\pi, -8) and (t, x) = \left(\frac{\pi}{4}, -1.5 \right), then the sinusoidal function is:

-8 +6.5\cdot \sin (-\pi\cdot \omega + \phi) = -8 (2)

-8+6.5\cdot \sin\left(\frac{\pi}{4}\cdot \omega + \phi \right) = -1.5 (3)

The resulting system is:

\sin (-\pi\cdot \omega + \phi) = 0 (2b)

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\frac{\pi}{4}\cdot \omega + \phi = \frac{\pi}{2} + 2\pi\cdot i, i \in \mathbb{Z} (3c)

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\pm \pi\cdot i + \pi\cdot \omega = \frac{\pi}{2} \pm 2\pi\cdot i -\frac{\pi}{4}\cdot \omega

\frac{3\pi}{4}\cdot \omega = \frac{\pi}{2}\pm \pi\cdot i

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To learn more on functions, we kindly invite to check this verified question: brainly.com/question/5245372

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Pleaseeee help ; will give brainliest ✨!!!
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