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Triss [41]
3 years ago
12

An inductor is connected to a voltage source and it is found that it has a time constant, t. When a 10-ohm resistor is placed in

series with the inductor, the time constant becomes t/2. A pure inductance of 30 mH is placed in series with the original inductor and the 10-ohm resistor, the time constant is t. What are the values of the original inductors (a) inductance, (b) internal resistance
Engineering
1 answer:
prisoha [69]3 years ago
6 0

Answer:

A) 30 mH

B ) 10-ohm

Explanation:

resistor = 10-ohm

Inductor = 30mH ( l )

L = inductance

R = resistance

r = internal resistance

values of the original Inductors

Note : inductor = constant time (t)  case 1

inductor + 10-ohm resistor connected in series = constant time ( t/2) case2

inductor + 10-ohm resistor + 30 mH inductance in series = constant time (t) case3

<em>From the above cases</em>

case 1 = time constant ( t ) = L / R

case 2 = Req = R + r hence time constant  t / 2  = L / R + r  therefore

t = \frac{2L}{R+r}

case 3 = Leq = L + l , Req = R + r .  constant time ( t )

hence Z = \frac{L + l}{R + r} = t

A) Inductance

To calculate inductance equate case 1 to case 3

\frac{L}{R} = \frac{L + l}{R + r} =   L / 10 = (L + 30) / ( 10 + 10 )

= 20 L = 10 L + 300 mH

   10L = 300 m H

therefore L = 30 mH

B ) The internal resistance

equate case 1 to case 2

\frac{L}{R} = \frac{2L}{R + r}

R + r = 2 R  therefore  ( r = R )   therefore internal resistance = 10-ohm

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Answer:

Complex power=84 W

Explanation:

using equation

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s=600^2/2z

s=84+29.4j

using s=P+jq

complex power=P=84 W

8 0
4 years ago
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The 1st part of the cell theory states that all things are made of cells. Why was
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Answer:

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Explanation:

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A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol
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Answer:

Explanation:

Considering the relation of the equilibrium vacancy concentration ;

nv/N = exp (-ΔHv/KT)

Where T is the temperature at which the vacancy sites are formed

K = Boltzmaan constant

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Rearranging the equation and expressing in term of the temperature and plugging the values given to get the temperature. The detailed steps is as shown in the attached file

5 0
3 years ago
When calculating a resolved shear stress within a single unit cell, the angle between the applied stress direction and the slip
givi [52]

Answer: 35.3 °

Explanation:

Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.

The simplest crystal structures are those that have present only a single atom at each lattice point.

body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms

With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:

[1 1 0] [1 1 1]

λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))

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6 0
3 years ago
A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

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P_4=50\ kPa

h_4=0.7(2304.7)+340.5

    =1953.83 \ KJ/Kg

and,

P_3=15 \ mPa

h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

At P_4=50 \ Kpa,

h_f=340.54

or,

V_f=0.001030 \ m^3/Kg

then,

⇒ h_2=340.54+0.001030(15\times 10^{3}-50)

        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

         =\frac{2610.8-1953.83}{2610.8-1836.26}

         =84.818 (%)

The thermal efficiency of cycle will be:

⇒ n_C=\frac{W_T-W_P}{2_{in}}

         =\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}

         =28.45 (%)  

4 0
3 years ago
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