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Korolek [52]
2 years ago
6

Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and

a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?
Engineering
1 answer:
Alborosie2 years ago
5 0

Answer:

V_{pp}=2V

Explanation:

Source Voltage V= 120V

Frequency f=60Hz

Peak output voltage Vp=15V

Peak Output Voltage with filter V_p'=14V

Generally the equation for Peak to peak voltage is mathematically given by

V_p'=V_p-\frac{V_{pp}}{2}

Therefore

V_{pp}=2(V_p-v_p')

V_{pp}=2(15-14)

V_{pp}=2V

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Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25
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By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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For a LED diode that has a= 632 nm, then the A1 is equal to:​
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