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Naya [18.7K]
3 years ago
7

A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto

r #2 on the second branch has a value of 330 ohms. Calculate the total power for this circuit.
Engineering
1 answer:
likoan [24]3 years ago
8 0

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

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The flow of power cannot be reversed since the slider could not move the worm gears. Since the input has one continuous tooth and the output has not teeth there is no gear ratio and no change in torque and speed.

8 0
3 years ago
Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
3 years ago
For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin
Nitella [24]
The correct answer is B
4 0
3 years ago
Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w
I am Lyosha [343]

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

6 0
3 years ago
Read 2 more answers
What is the formula used to find the volume of this shape
horsena [70]
Volume=Hh(b1+b2)/2. b1 and b2 are the base of the trapezoid
8 0
4 years ago
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