1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Oxana [17]
3 years ago
7

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

principle in question is indeed correct.
Principles of Series Circuits:
The voltages across each circuit element add to that of the battery.
The current through each circuit element is the same.
Higher resistances have higher voltage across them.
Resistors added in series to any circuit increase the resistance to current in the circuit.

Principles of Parallel Circuits:
Voltages across each circuit element are the same.
The current through each parallel circuit element adds to that going through the battery.
The higher resistance in a parallel circuit has less current.
Resistors added in parallel to any circuit reduce the resistance to current in the circuit.

General Principles:
A branch in a circuit that contains a short has no voltage across it.
A branch in a circuit that is open has all the voltage across it.
A battery is a constant voltage source.
A battery is not a constant current source.
Ammeters have very little internal resistance.
Voltmeters have very large internal resistance.

Engineering
1 answer:
Nata [24]3 years ago
3 0

<u>Explanation</u>:

For series

\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)

\(I=I_{1}=I_{2}=I_{n}\) (current is the same)

V=I R(\text {voltage is directly proportional to } R)

R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }

For parallel

\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })

I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})

\(I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)\)

\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}

You might be interested in
4. Two technicians are discussing the evaporative emission monitor. Technician A says that serious monitor faults cause a blinki
snow_lady [41]

Answer:

The correct option is;

Neither Technician A nor B

Explanation:

The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion

It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars

Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown

It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.

Therefore, the correct option is that neither Technician A nor B are correct.

3 0
3 years ago
Sarah is developing a Risk Assessment for her organization. She is asking each department head how long can they be without thei
Natali [406]

Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

8 0
3 years ago
What camera battery does the canon 250d use?
Mila [183]

Answer:

The Canon EOS 250 is a digital DSLR that features a 24MP sensor. It is powered by the Canon LP-E17, which is a rechargeable Lithium-Ion battery pack.

Explanation:

8 0
3 years ago
Read 2 more answers
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
gladu [14]

Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

4 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
andre [41]

Answer:

4.17x10^-3 kW/K

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
3 years ago
Other questions:
  • From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye
    12·1 answer
  • What is pessimism technology
    12·1 answer
  • The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm
    5·1 answer
  • A banked highway is designed for traffic moving at v = 88 km/h. The radius of the curve r = 314 m. show answer No Attempt 50% Pa
    5·2 answers
  • What fraction of the worlds surface is estimated to be arable land?
    15·1 answer
  • A preheater involves the use of condensing steam at 100o C on the inside of a bank of tubes to heat air that enters at I atm and
    14·1 answer
  • Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.
    12·1 answer
  • The ultimate BOD of a river just below a sewage outfall is 50.0 mg/L, and the oxygen deficit at the outfall D0 is 2.0 mg/L. The
    6·1 answer
  • What did Congress do in 1787 to settle land disputes among the settlers?
    11·1 answer
  • I need to solve for d
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!