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3 years ago

5. What are the 3 basic types of electrical circuits?

Engineering

1 answer:

myrzilka [38]3 years ago

4 0

Answer:

D. Series, Parallel, and Series/Parallel

Explanation:

Categorized according to topology, circuits are ...

Series, Parallel, and Series/Parallel

Circuits can also be categorized according to the nature of the voltages and currents found in them. These categories would include AC, DC, and some mixture of AC and DC.

The terms "simple" and "direct" and "indirect" don't seem to have any defined meaning in this context.

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An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s

nignag [31]

Given:

$X_{L} = 50.0 \ohm$

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

$V_rms} = 85.0 V$

Solution:

To calculate max current in inductor, $I_{L(max)$ :

At f = 60.0 Hz

$X_{L} = 2\pi fL$

$50.0 = 2\pi\times 60.0\times L$

L = 0.1326 H

Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

Now, $I_{L(max)$ is given by:

$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

$I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A$

Therefore, max current in the inductor, $I_{L(max)$ = 2.13 A

7 0

3 years ago

Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for

lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0

3 years ago

An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with

Alex_Xolod [135]

An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with a blue stripe or dot.

8 0

2 years ago

Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No

Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is $1.557\times10^{-7}$ m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

$Re=\frac{vd}{\nu}$

Here, v is velocity, $\nu$ is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

$Re=\frac{vd}{\nu}$

$Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}$

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0

3 years ago

A single-cylinder pump feeds a boiler through a delivery

Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 $m^3/hr$

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius $r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m$

We know that area is given by

$A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2$

We know that discharge is given by $Q=AV$ , here A is area and V is velocity

So $Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec$

So net discharge in 1 hour = $981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour$

8 0

3 years ago