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professor190 [17]
3 years ago
14

Identify the true statements about checking resistors with an ohmmeter. (Check all that apply.) Multiple select question. When c

hecking resistance in a circuit, there should not be any parallel resistance paths. When checking resistance in a circuit, there should be parallel resistance paths present. The ohmmeter leads are connected across the resistance to be measured. An ohmmeter has its own voltage source.
Engineering
1 answer:
ipn [44]3 years ago
5 0

Answer:

Please find the answer in the explanation.

Explanation: To Identify the true statements about checking resistors with an ohmmeter,

When checking resistance in a circuit, there should be parallel resistance paths present. Because any circuit or resistors to be measured has to be shunted. That is connecting it in parallel so that small amounts of current can pass through the circuit.

The ohmmeter leads are connected across the resistance to be measured. An ohmmeter has its own voltage source.

The ohmmeter makes use of batteries as power source.

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Explanation:

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Answer: Power

Explanation:

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Read 2 more answers
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

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We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

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4 years ago
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