a 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?

Answers

Answer 1

Answer:

No.

Explanation:

[tex]P_r[/tex] = Power rating = 1 W

R = Resistance = [tex]350\ \Omega[/tex]

V = Voltage = [tex]24\ \text{V}[/tex]

Power is given by

[tex]P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}[/tex]

[tex]1.65\ \text{W}>1\ \text{W}[/tex]

So

[tex]P>P_r[/tex]

Hence, the resistor is not operating within its power rating.


Related Questions

Water vapor at a rate of 40,000 kg/h and at 8 MPa and 500 C enters an adiabatic turbine and leaves the turbine at 40 kPa as saturated vapor. The rate of entropy generation for this turbine is close to,

Answers

Answer:

M(dot)= 40000 kg/h Pressure= 8Mpa and T= 500C If its adiabatic then you know that it loses no heat or Q(dot) =0 You have a turbine which drives a shaft producing work for the other systems.We want to find entropy production of alpha.

An alternating current is supplied to an electronic component with a rating that it be used only for voltages below 12 V. What is the highest Vrms that can be supplied to this component while staying below the voltage limit

Answers

Answer:

[tex]V_{rms}=6\sqrt2\ V[/tex]

Explanation:

Given that,

The maximum voltage of an alternating current, [tex]V_{max}=12\ V[/tex]

We need to find the highest Vrms that can be supplied to this component while staying below the voltage limit.

Let rms voltage in terms of peak voltage is given by :

[tex]V_{rms}=\dfrac{V}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\times \dfrac{\sqrt2}{\sqrt2}\\\\=\dfrac{12\sqrt2}{2}\\\\=6\sqrt2\ V[/tex]

Hence, the required rms voltage is [tex]6\sqrt2\ V[/tex].

What’s are the preventive strategies for workplace violence

Answers

Adopt a formal workplace violence prevention training policy and program, and communicate it to employees. Have managers take an active role in employee awareness of the plan; make sure they are alert to warning signs of workplace violence and know how to respond

who designs each page or screen with which a user interacts?​

Answers

Answer:

UI Design

Explanation:

A fluid flows though a horizontal 0.1 inch diameter pipe. When the Reynold number is 1508, the head loss over a 20-ft length of the pipe is 6.2 ft. Determine the fluid velocity.

Answers

Answer:

The fluid velocity V = 1.98 ft/s

Explanation:

From the information given:

The fluid velocity can be determined from the head-loss [tex]h_L[/tex] of a laminar pipe and it is expressed as:

[tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]

where;

f = frictional factor ; l = length; D = diameter; V= fluid velocity and g = acceleration due to gravity.

And;

[tex]f = \dfrac{64}{Re}[/tex]

For fluid movement in a laminar flow, the Reynolds number (Re) is usually lesser than 2100.

Given that:

Re = 1508 < 2100   ( laminar flow)

Then;

[tex]f = \dfrac{64}{1508}[/tex]

f = 0.04244

Also;

the head-loss [tex]h_L[/tex] = 6.2 ft

frictional force f = 0.04244

length = 20-ft

acceleration due to gravity (g) = 32.2 ft/s²

Replacing all the values into the equation [tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]; the fluid velocity is:

[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.1 \times \dfrac{1}{12} \times 2\times 32.2}[/tex]

[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.53667}[/tex]

6.2 × 0.53667 = 0.04244 × 20 × V²

3.327354  = 0.8488 × V²

[tex]V^2= \dfrac{3.327354} { 0.8488}[/tex]

[tex]V^2=3.92[/tex]

[tex]V = \sqrt{3.92}[/tex]

V = 1.98 ft/s

A piston–cylinder device contains 15 kg of saturated refrigerant-134a vapor at 280kPa. A resistance heater inside the cylinder with a value of 110V is used to transfer heat of 20.67kJ/kg to the refrigerant at constant pressure for 6min. Determine the current supplied if the final temperature is 75℃

Answers

Answer:

18.77 A

Explanation:

To solve this we use the energy balance equation, that is:

[tex]E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\ T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\[/tex]

[tex]39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A[/tex]

Type the correct answer in the box. Spell all words correctly.

A genetically engineered hormone, , can treat Mary’s child for growth hormone deficiencies by stimulating body growth and increasing muscle mass.

Answers

Answer:

recombinant human growth

Explanation:

Answer:

recombinant human growth

Explanation:

yes

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