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liq [111]
3 years ago
10

Explain how you would solve for total resistance in a parallel circuit versus a series circuit. How would you apply and solve fo

r Kirchhoff’s current law in a parallel circuit, assuming you know the values of each resistor?
Engineering
1 answer:
Mekhanik [1.2K]3 years ago
8 0

Answer:

No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.

Explanation:

You might be interested in
Draw a FBD of the beam with reactions at A & B. A is a pin, B is a roller. Try to guess intuitively which way the vertical c
Vesnalui [34]

Answer:

kindly check the drawing of the  FBD of the beam with reactions at A & B. A is a pin, B is a roller in the attached picture.

Explanation:

Without further ado, let's dive straight into the solution to the question above. From the diagram of the FBD of the beam with reactions at A & B it can be shown that the reaction moment is anticlockwise while the moment is clockwise.

The system is at equilibrium and the it does not matter where you place the couple (pure) moment.

The distance from A to C can either be equal or not. If AY = 2.15 kN and M = 25.8. Then, the distance between A and B = 25.8/2.15 = 12m.

5 0
3 years ago
For a 3-Phase, Wye connected system the Line to Line Voltage measures 12,470 Volts, the Phase current measures 120 Amps.
vladimir2022 [97]

Answer:

A. 7199.55 volts

B. 120A

Explanation:

In this question we have the

line voltage = VLL = 12470volts

Phase current = Iph = 120 amps

A.)

We are to calculate the line-to-neutral/phase voltage here

VLL = √3VL-N

VL-N = VLL/√3

VL-N = 12470/√3

This gives a line to neutral phase/voltage of 7199.55 volts.

B.

We are to calculate the line current here:

In this connection, the line current and the phase current are equal

ILL = Iph = 120A

6 0
3 years ago
The consumer price index (CPI) for a given year is the amount of money in that year that has the same purchasing power as $100 i
Kipish [7]

Answer:

The formula should be 234.1 x (1 + 2.9%)^t

Explanation:

As the questions indicates, the CPI indicates what the value for money is in that given year compared to the base year. So a higher CPI amount is given as a result of inflation over the years. So the rate of inflation was such that the 100 dollars in 1983 (taken as the base year) is equivalent to 234.1 in 2015 as a result of inflation. So, essentially, the time value of money concept is being applied here. We can use the formula for calculating the future value of money over here as well.

The equation is: FV = PV × (1 + x%)^t

where, FV is future value, PV is the present value today, x% is the rate of change, and t is the time period in years.

With the values at hand, the formula to compute what the CPI should be t years after 2015,

CPI (t years after 2015) = 234.1 x (1.029)^t

3 0
4 years ago
The bolts that attach a bracket to an industrial machine must each carry a static tensile load of 4 kN.
HACTEHA [7]

Answer:

M10 × 1.5

least number of threads is 4.3

Explanation:

given data

static tensile load = 4 kN = 4000 N

safety factor = 5

coarse‐thread metric = 5.8

solution

we know that proof load for 5.8  class is Sp = 380 MPa

so area of cross section is express as

area = \frac{force \times FOS}{Sp}    ......................1

area = \frac{4000 \times }{380}

area = 52.6  mm²

so by table at 58 mm² we can say we resist  M10 × 1.5

and

bolt tensile strength is express as

bolt tensile strength = nut shear strength    ......2

so here bolt tensile strength = At × Sy (bolt)    ...............3

and nut shear strength =  πd ( 0.75 t ) Sys

nut shear strength =   π × 10 × ( 0.75 t )  × 0.58  × \frac{2}{3}   × Sy(bolt)   ............4

so from equation 3 and 4 we get

t = 6.37 mm

and

for the pitch is = 1.5 mm

least number of threads is 4.3

8 0
3 years ago
Exercise 5.Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressureof 275 kPa at an elevation of 2
wariber [46]

Answer:

a. Pressure head: 33.03,

Velocity Head: 6.53

b. Pressure Head: -1.97,

Velocity Head: 6.53

Explanation:

a.

Given

Diameter = 0.15-m, radius = 0.075

rate = 0.2 m3/s

Pressure =275 kPa

elevation =25 m.

We'll consider 3 points as the water flow through the pipe

1. At the entrance

2. Inside the pipe

3. At the exit

At (1), the velocity can be found using continuity equation.

V1 = ∆V/A

Where A = Area = πr² = π(0.075)² = 0.017678571428571m²

V1 = 0.2/0.017678571428571

V1 = 11.32 m/s

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:

P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2

Substitute in the values

P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25

P1/yH20 = (275 * 10³Pa)/yH20 + 25 - 25

=> P1/yH20 = (275/9.81 + 5)

P1/yH20 = 33.03

The velocity head at point one is then given by

V2²/2g = 11.32²/2 * 9.8

V2²/2g = 6.53

b.

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:

P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

P3/yH20 = 33.03 + 20 - 55

=> P1/yH20 = -1.97

The velocity head at point three is then given by

V2²/2g = V3²/2g = 6.53

4 0
3 years ago
Read 2 more answers
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