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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}}

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In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

3 0
3 years ago
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
3 years ago
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