An inverting amplifier with +11 V supply voltages normally has a sinusoidal output of 10 Vpp. When checking the circuit with an
oscilloscope, you find that the output is 0 V. Which of the following could account for this problem? A) RF is shorted by a solder bridge.B) R1 is open. C) Vi = 0 D) All of the above
1 answer:
Answer:
D) All of the above
Explanation:
To understand why we need to study case by case:
A) RF is shorted by a solder bridge: If RF is shorted, the output is going to be at the same potential that the inverting input due to a virtual short circuit (the inverting input is 0V because the non-inverting input is GND)
B) R1 is open: If R1 is open there is not input voltage to amplify therefore the output is 0V
C) Vi = 0: If Vi is equal zero there is not input voltage to amplify therefore the output is 0V
Note: Because there's is not schematic we assume the one in the picture down below, our logical explanation is complemented with a simulation that matches our results. For our case the amplifier requires a positive and negative supply, so we use an inverting amplifier with +11 V and -11V power supply.
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Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
Answer:
Explanation: see attachment below
