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.
You might be interested in
Answer:
Given:
high temperature reservoir
low temperature reservoir
thermal efficiency
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as
The thermal efficiency of second heat engine can be written as
The temperature of intermediate reservoir can be defined as
Answer:
33.56 ft^3/sec.in
Explanation:
Duration = 6 hours
drainage area = 185 mi^2
constant baseflow = 550 cfs
<u>Derive the unit hydrograph using the inverse procedure </u>
first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below
Vdrh = sum of drh * duration
= 29700 * 6 hours ( 216000 secs )
= 641,520,000 ft^3.
next step : Calculate the volume of runoff in equivalent depth
Vdrh / Area = 641,520,000 / 185 mi^2
= 1.49 in
Finally derive the unit hydrograph
Unit of hydrograph = drh / volume of runoff in equivalent depth
= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in
