While discussing the diagnosis of an EI system in which the crankshaft and camshaft sensor tests are satisfactory but a spark te
1 answer:
Answer:
Both A and B are correct.
Explanation:
The crankshaft sensor is attached to the crankshaft pulley on the
back side of the crankshaft. It is used to pass a signal to the PCM
which controls the ignition of an engine. This sensor helps in providing some basic timing for the signal. If there is a faulty sensor present, the signal will not be received by the PCM. So, it will affect the engine while starting. One reason can be a defective
assembly of the coil.
If there is a faulty , the signal will not be received properly. This will not allow the passing of signals to the ignition system. This will cause a misfire. One of the most common causes of
misfire is a faulty ignition coil. This will affect the working of a
sensor. So, prevent engine from starting.
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Answer:
The minimum mass flow rate will be "330 kg/s".
Explanation:
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For steam,
For water,
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⇒
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complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
