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irakobra [83]
3 years ago
14

An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100

Q. drives a load of 500 Q. What voltage and power gains (expressed in dB) would you expect with the load connected
Engineering
1 answer:
Vanyuwa [196]3 years ago
3 0

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

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EastWind [94]

Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

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ANSWER: a=> True.

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8 0
3 years ago
find magnitude of the resultant force, if 30N,40N,50N and 60N forces are acting along the line joining the centr of a square to
viktelen [127]

Answer:

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7 0
3 years ago
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:

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For refrigeration

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

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

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

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

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So

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

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For heat pump

So COP of heat pump is given as follows

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

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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

5 0
3 years ago
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erik [133]

Answer:

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