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.
Answer:
¿Qué aplicación estás usando? podría ser porque te equivocaste un paso
Answer:
#include <stdio.h>
void SplitIntoTensOnes(int* tensDigit, int* onesDigit, int DecVal){
*tensDigit = (DecVal / 10) % 10;
*onesDigit = DecVal % 10;
return;
}
int main(void) {
int tensPlace = 0;
int onesPlace = 0;
int userInt = 0;
userInt = 41;
SplitIntoTensOnes(&tensPlace, &onesPlace, userInt);
printf("tensPlace = %d, onesPlace = %d\n", tensPlace, onesPlace);
return 0;
}
The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.
<h3>What are the benefits of practicing safety in the laboratory?</h3>
A laboratory is known to be one that is known to have a lot of potential risks that is said to often arise due to a person's exposure to chemicals that are corrosive and toxic, flammable solvents, high pressure gases and others.
Therefore, A little care and working in line to all the prescribed safety guidelines will help a person to be able to avoid laboratory mishaps.
Therefore, The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.
Learn more about safety measures from
brainly.com/question/26264740
#SPJ1
Answer:
At 3 =99.60
At 6 =66.66
At 12 =33.38
Please see attachment for the percent consolidation.
