how to solve circuit theory using mesh analysis
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Answer:

I think ( MOV Metal oxide varistors )

Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more MOV Metal oxide varistors than bypass and absorb the energy of the transient.

Answer:

Digestion functions become more active

Explanation:

I just took the text!

Answer:

[tex]V_z=9.1v[/tex]

[tex]V_{zo}=8.74V[/tex]

[tex]I=10mA[/tex]

[tex]R=589 ohms[/tex]

Explanation:

From the question we are told that:

Zener diode Voltage [tex]V_z=9.1-V[/tex]

Zener diode Current [tex]I_z=9 .A[/tex]

Note

[tex]rz = 40\\\\IZK= 0.5 mA[/tex]

Supply Voltage [tex]V_s=15[/tex]

Reduction Percentage [tex]P_r= 50 \%[/tex]

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

[tex]V_z=V_{zo}+I_zr_z[/tex]

[tex]9.1=V_{z0}+9*10^{-3}(40)[/tex]

[tex]V_{zo}=8.74V[/tex]

Therefore

[tex]At I_z-10mA[/tex]

[tex]V_z=V_{z0}+I_zr_z[/tex]

[tex]V_z=8.74+(10*10^{-3}) (40)[/tex]

[tex]V_z=9.1v[/tex]

Generally the equation for Kirchhoff's Current Law is mathematically given by

[tex]-I+I_z+I_l=0[/tex]

[tex]I=10mA+\frac{V_z}{R_l}[/tex]

[tex]I=10mA+\frac{9.1}{0}[/tex]

[tex]I=10mA[/tex]

Therefore

[tex]R=\frac{15V-V_z}{I}[/tex]

[tex]R=\frac{15-9.1}{10*10^{-3}}[/tex]

[tex]R=589 ohms[/tex]

Answer:

320 m

Explanation:

To find the side length of the current park, x, we solve the quadratic equation for the area of the park

x² + 200x = 166400

x² + 200x - 166400 = 0

We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x

So, we re-write the expression as

x² + 200x - 166400 = 0

x² + 520x - 320x - 166400 = 0

We write out the factors of the expression,

x² + 520x - 320x - 320 × 520 = 0

Factorizing the expression, we have

x(x + 520) - 320(x + 520) = 0

(x + 520)(x - 320) = 0

x + 520 = 0 or x - 320 = 0

x = -520 or x = 320

Since x is not negative, we take the positive answer.

So, x = 320 m

Answer:

A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:

Answer:

[tex]l=24mm[/tex]

Explanation:

From the question we are told that:

Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]

Y value [tex]Y=1.0[/tex]

Stress level of[tex]\sigma =200 MPa[/tex]

Generally the equation for length of a surface crack is mathematically given by

[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]

[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]

[tex]l=0.024m[/tex]

Therefore

in mm

[tex]l=24mm[/tex]

Answer:

ROLAP  is a branch of OLAP that is used to contain Relational database ( RDB ). which is a very fast database ( quick process of queries )

Explanation:

ROLAP ( Relational On-line Analytical processing ) is a branch of OLAP that is used to contain Relational database ( RDB ).

Advantages of ROLAP ( reasons for the use of ROLAP )

Answer:

The answer is below.

Explanation:

Some of the ways, how I think elliptic cryptography is more advanced than RSA are the following:

1. ECC - Elliptic Curve Cryptography uses smaller keys for the same level of security, particularly at greater levels of security.

2. ECC can work well and at a faster rate on a small-capacity device compared to RSA

3. It uses offer speedier SSL handshakes that enhance security

4. It offers fast signatures

5. It allows signatures to be computed in two stages, which enables lower latency than inverse throughput.

6. Relatively quick encryption and decryption

Answer:

Explanation:

To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...

  240/12 = 20 . . . batteries needed

__

The current draw will be ...

  (3000 W)/(240 V) = 12.5 A

Then the time available from the battery stack is ...

  (120 Ah)/(12.5 A) = 9.6 h

The motor can run 9.6 hours from the series connection.

Answer:

Lack of oxygen

Decompression sickness

temperature variation

lack of gravity

Cosmic radiations hazards

motion sickness

Explanation:

When an astronaut travels in the space he is aware of the challenges he might face during his journey. There are various test and rehearsals of travelling before a final travel takes place. An astronaut goes through many challenges which includes lack of oxygen supply, decompression and motion sickness. There is no gravity in space so an astronaut will have to be aware of the difficulties he might face during his travel. An hour on earth is 7 years long in space, so an astronaut should have patience and be able to deal with time variation.

Answer: (b)

Explanation:

Given

Original length of the rod is [tex]L=100\ cm[/tex]

Strain experienced is [tex]\epsilon=82\%=0.82[/tex]

Strain is the ratio of the change in length to the original length

[tex]\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm[/tex]

Therefore, new length is given by (Considering the load is tensile in nature)

[tex]\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm[/tex]

Thus, option (b) is correct.

Answer:

very good thx

Explanation:

Answer:

a) 1.918 kw

b) 86.23%

c) 0.26 kw

Explanation:

Given data:

T1 = -30°C = 243 k  , T0 = 27°C

using steam tables

h1 = 232.19 KJ/kg

s1 = 0.9559 Kj/Kgk

T2 = 55°C   P2 = 900 kPa

Psat = 1492 kPa,  h2 = 289.95 Kj/Kg, s2 = 0.9819 Kj/kgk , m = 0.0332 kg/s

a) Determine the power input to the compressor

power input = 1.918 kw

b) Determine isentropic efficiency of compressor

Isentropic efficiency = 86.23%

c) Determine rate of exergy destruction

rate = 0.26 kw

Attached below is the detailed solution of the given problems

Answer: Hello your question is incomplete attached below is the complete question

answer:

attached below

Explanation:

In this Verilog description we will refer to figure attached below

we will make some representation which are :

Represent  outputs of the input  AND gates = P

Represent outputs of the input NOR gates = Q

Inverter = R

attached below is the Verilog description

Answer:

Technician B only.

Explanation:

It is not necessary that all electric motors will use Direct current, some may also use Alternative current. Some electric motors that use DC may use brushes. There are two types of brushless motors that use alternative current.

Answer:

[tex]V_{pp}=2V[/tex]

Explanation:

Source Voltage [tex]V= 120V[/tex]

Frequency [tex]f=60Hz[/tex]

Peak output voltage [tex]Vp=15V[/tex]

Peak Output Voltage with filter [tex]V_p'=14V[/tex]

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

[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]

Therefore

[tex]V_{pp}=2(V_p-v_p')[/tex]

[tex]V_{pp}=2(15-14)[/tex]

[tex]V_{pp}=2V[/tex]

Answer:

mark me as a brainleast

Explanation:

209781

Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

[tex]t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}[/tex]

t = 6179.1 s = 102.9 min = 1.7 h

i have made notes and saved it as a pdf u can take it to answer question and make ur concept good

The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m

Stopping sight distance; S = 220 m

Design Speed; V = 110 km/h

Intersection grade 1; G1 = +2.7

Intersection Grade 2; G2 = -3.5

From the table, at S = 220 m and V = 110 km/h, we can see that;

Radius of vertical curvature; K = 74

A = G1 - G2

A = 2.7 - (-3.5)

A = 2.7 + 3.5

A = 6.2

L = KA

Thus;

L = 74 × 6.2

L = 458.8 m

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Answer:

i) when circuit is unloaded : R1 + R2 = 2kΩ.

ii) when 5V output voltage is applied :  R1 = 1 kΩ , R2 =  1 kΩ

iii) when 2.5 v output voltage is applied : R1 = 1500 Ω,  R2 = 500 Ω

iv) when: R1 = 1 kΩ , R2 =  1 kΩ  is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

Explanation:

Current drawn from source under loaded condition ≤ 5 mA

source voltage = 10 v ,  required output = 5 v , 2.5 v

attached below is the sketch of the circuit

Resistance values

i) when the circuit is unloaded

Req = R1 + R2 = 2 kΩ  ( Req = Vs / I  = 10 / 5*10^-3 = 2 kΩ )

ii) when  output voltage = 5 v

we will apply voltage divider rule

R1 = 1 kΩ ,

R2 =  1 kΩ

iii) When the output voltage = 2.5 v

applying voltage divider rule

R1 = 1500 Ω

R2 = 500 Ω

iv) when the load is connected to each tap one at a time

i.e. when the resistance are in parallel

when: R1 = 1 kΩ , R2 =  1 kΩ  is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

attached below is the detailed solution to the given problem

Answer:

Required pump power = - 11.205 kW

Explanation:

Pi = 600 kPa

Zi ( initial height ) = - 5m

P2 = 200 kPa

Ze ( top floor above ground level ) = 150m

Flowrate = 0.01 m^3/s = 10 kg/s

Temperature = 10°C  ( assumed value )  remains constant

calculate the value of pumping power required ( i.e. work done )

Applying the energy equation

Hi + 1/2(vi)^2 + gZi = He + 1/2(Ve)^2 + gZe + W

given negligible internal/kinetic  energy  difference in "h's = the pv terms

W = ( Pi Vi - Pe V2 ) + g( Zi - Z2 )

   = ( (600 * 0.001)  - (200 * 0.001) ) + (9.81/1000) ( - 5 - 150 )

   = - 1.1205 Kg/kJ

required pump power = flowrate * -1.1205

                                     = 10 * ( - 1.1205 ) = - 11.205 kW

Answer:

  your answer is correct

Explanation:

You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...

 A B Y

 0 0 1

 0 1 0

 1 0 0

 1 1 1

That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.

Answer:

A) 600.8 m/s

B) (i) 22.5 cm^2   (ii) 7.11 cm^2

Explanation:

Given data :

P1 = 20 bar ,  T1 = 2808°C

P2 = 7 bar ,  T2 = 1808°C

mass flow rate = 1.5 kg/s

Using the superheated vapor region in Table A-4

h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg

h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg  

A) calculate exit velocity  ( m/s )

given that we are to neglect heat transfer and potential energy

V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )

                             = [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )

                             = 600.8 m/s

B) calculate the inlet and exit flow areas ( cm^2 )

i) Inlet flow area

A1 = ( m * v1 ) / V1

    = ( 1.5 * 0.1200 )/ 80  = 22.5 cm^2

ii) exit flow area

A2 = ( m * v2 ) / V2

    = ( 1.5 * 0.2847 ) / 600.8  = 7.11 cm^2

Explanation:

print() has a variable number of parameters. this is the answer.

hope this helps you

have a nice day

Answer:

The correct answer is "99.9%".

Explanation:

According to the information given in the question,

[tex]P(1 \ fail) = 0.1[/tex]

The probability of all fail will be:

[tex]P(all \ fail) = (0.1)^3[/tex]

                  [tex]=0.001[/tex]

hence,

[tex]P(not \ all \ fail)= 1-P(all \ fail)[/tex]

                        [tex]=0.999[/tex]

                        [tex]=99.9[/tex] (%)

Thus the above is the right answer.

Answer:

a) [tex]\mu=3.3[/tex]

b)  [tex]Q=1440.7KJ/Kmol[/tex]

c)  [tex]W=1872.9KJ/Kmol[/tex]

Explanation:

From the question we are told that:

Initial Temperature [tex]T_1=325k[/tex]

initial Pressure [tex]P_1=2 bar[/tex]

Final Pressure [tex]P_2=4 bar[/tex]

iso-thermal expansion [tex]T_2=250k[/tex]

a)

Generally the equation for Coefficient of performance is mathematically given by

[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]

[tex]\mu=\frac{250}{325-250}[/tex]

[tex]\mu=3.3[/tex]

b)

Generally the equation for Heat Expansion is mathematically given by

[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]

Where

R=Gas constant

[tex]R=8.314462618[/tex]

Therefore

[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]

[tex]Q=1440.7KJ/Kmol[/tex]

c)

Generally the equation for work input is mathematically given by

[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]

[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]

[tex]W=1872.9KJ/Kmol[/tex]

The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol

Given Data:

Isothermal expansion occurs at 250K.

This is calculated as

COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]

[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]

Therefore; In isothermal process du = 0

R = 8.314 AkJ/K.mol

Q = 8.314 * 250 In(4/2)

Q = 1440.71kJ/K.mol

W[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]

The work input is 1872.92kJ/K.mol

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