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3 years ago

1. Examine the following circuit. Find RT, I3, R1, R2, R3, V1, V2 and V3. Show all of your work clearly below.

Engineering

1 answer:

Mkey [24]3 years ago

6 0

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

R2 = (8 V)/(4 A) = 2 Ω

R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω

V1 = V2 = V3 = 8 V

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 k

ad-work [718]

Answer:

See attached image for diagrams and solution

6 0

3 years ago

True or false? if i were to hook up an ac voltage source to a resistor, the voltage drop across the resistor would be in phase w

hodyreva [135]

Answer: True

Explanation:

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2 years ago

A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can

Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a) ln(T2/T1)=μβ β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb T1=80 lb μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0

3 years ago

In Lab 7, we worked through a program that displayed the homeless shelter occupancy over time. The same approach can be used for

Bezzdna [24]

Answer:

Explanation:

The python code to generate this is quite simple to run.

i hope you understand everything written here, you can as well try out other problems to understand better.

First to begin, we import the package;

Code:

import pandas as pd

import matplotlib.pyplot as plt

name = input('Enter name of the file: ')

op = input('Enter name of output file: ')

df = pd.read_csv(name)

df['Date'] = pd.to_datetime(df["Date"].apply(str))

plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")

plt.legend(loc="upper right")

plt.xticks(rotation=20)

plt.savefig(op)

plt.show()

This should generate the data(plot) as seen in the uploaded screenshot.

thanks i hope this helps!!!

6 0

3 years ago