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Jlenok [28]
3 years ago
9

Calculate the impedance of a 20 mH inductor at a frequency of 100 radians/s. Calculate the impedance of a 500 µF inductor at a f

requency of 100 radians/s. What is the equivalent impedance of the inductor and capacitor connected in series? What is the equivalent impedance of the inductor and capacitor connected in parallel?
Engineering
2 answers:
Naily [24]3 years ago
7 0

Given Information:  

Capacitance = C = 500 μF

Inductance = L = 20 mH

Frequency = ω = 100 rad/sec

Required Information:  

Capacitive reactance = Xc = ?

Inductive reactance = XL = ?

Equivalent series impedance = Zseries = ?

Equivalent parallel impedance = Zparallel = ?

Answer:

Capacitive reactance = Xc = -j20 Ω  

Inductive reactance = XL = j2 Ω

Equivalent series impedance = Zseries = -j18 Ω

Equivalent parallel impedance = Zparallel = j2.22 Ω

Explanation:

The capacitive reactance is calculated using

Xc = 1/jω*C

Where ω is the frequency in rad/sec and C is the capacitance of the capacitor

Xc = 1/j100*500x10⁻⁶

Xc = -j20 Ω

The inductive reactance is calculated using

XL = jω*L

Where ω is the frequency in rad/sec and L is the inductance of the inductor

XL = j100*0.02

XL = j2 Ω

The equivalent series impedance an can be calculated as

Zseries = XL + Xc

Zseries = j2 - j20 Ω

Zseries = -j18 Ω

The equivalent parallel impedance an can be calculated as

1/Zparallel = 1/XL + 1/Xc

Simplifying the above equation yields,

Zparallel = (XL*Xc)  / (XL+ Xc)

Zparallel = (j2*-j20)/(j2-j20)     ( j*j = -1)

Zparallel = 40/-j18     (1/-j = j)

Zparallel = j2.22 Ω

Helga [31]3 years ago
3 0

Answer:

a) Zinductor = j2 Ohm

b) Zcapacitor = -j20 Ohm

c) Zseries = -j18 Ohm

d) Zparallel = -j2.22 Ohm

Explanation:

The inductor impedance is directly proportional to the frequency, therefore:

a) frequency = 100 rad/s inductor = 20mH

Zinductor = j*frequency*inductor = j*100*20*10^(-3) = j2 Ohm

b) The capacitor inductance is inversely proportional to the frequency and it's also negative, therefore:

frequency = 100 rad/s capacitor = 500 uF

Zcapacitor = j*frequency*capacitor = -j/(100*500*10^(-6)) = -j/(0.05) = -j20 Ohm

c) The equivalent impedance of these two components in series is the sum of each part, wich goes as follow:

Zseries = Zinductor + Zcapacitor = j2 -j20 = -j18 Ohm

d) The equivalent impedance of these two components in parallel is given by the following equation:

Zparallel = (Zinductor*Zcapacitor)/(Zinductor+Zcapacitor)

Zparallel = [j2*(-j20)]/(j2-j20)

Zparallel = [-40]/(-j18) = 2.22/j = -j2.22 Ohm

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Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

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Assumptions for the calculation are as follows

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During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

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-k\frac{dT }{dx } = hc (T - T∞) at point x = L

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From the second integration we have

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From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

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