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ira [324]
2 years ago
9

Find the capacitance reactance of a 0.1 micro frequency capacitor 50Hz and at 200Hz​

Engineering
1 answer:
Paraphin [41]2 years ago
5 0

Complete Question:

Find the capacitive reactance of a 0.1 microfarad capacitor with frequency 50Hz and at 200Hz​.

Answer:

I. Capacitive reactance = 31826.86 Ohms.

II. Capacitive reactance = 7956.72 Ohms.

Explanation:

<u>Given the following data;</u>

Capacitance = 0.1 uF = 0.0000001 Farad

Frequency = 50 Hz and 200 Hz

To find the capacitive reactance;

Mathematically, the capacitive reactance of an electronic circuit is given by the formula;

X_{c} = \frac {1}{2 \pi fc}

Where;

  • Xc is capacitive reactance.
  • f is the frequency.
  • c is the capacitance.

Substituting into the formula, we have;

I. At frequency, f = 50 Hz

X_{c} = \frac {1}{2 * 3.142 * 50 * 0.0000001}

X_{c} = \frac {1}{314.2 * 0.0000001}

X_{c} = \frac {1}{0.00003142}

X_{c} = 31826.86

II. At frequency, f = 200 Hz

X_{c} = \frac {1}{2 * 3.142 * 200 * 0.0000001}

X_{c} = \frac {1}{1256.8 * 0.0000001}

X_{c} = \frac {1}{0.00012568}

X_{c} = 7956.72

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                        \frac{lo*lo*lo}{l1*l2*l3}=1.0

                      taking natural log on both sides

                              ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)

Considering the logarithmic Laws of division and multiplication :

                                ln(AB) = ln(A)+ln(B)

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                           ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0

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8 0
3 years ago
If you had a match and a lantern and a candle in the dark which one would you choose to light.
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Answer:

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Explanation:

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I don't know of this is a homework question, but I answered it anyway :)

5 0
3 years ago
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2 years ago
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

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2 years ago
The 5-kg collar has a velocity of 5 m&gt;s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

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lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

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Replacing in eq. 1:

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Replacing values and clearing vB:

vB = 5.33 m/s

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Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

6 0
3 years ago
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