A and C is the answer to the question. Be 15 years old & get a permit
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.
The precautions that should be taken to avoid the overloading of domestic electric circuits are:
- Do not put high voltage wires in one socket.
- Do not use many electric appliances of high power at the same time.
<h3>What are electric circuits?</h3>
Electric circuits are wires or devices that give electricity to devices that run on electricity. Running of electric devices should be done carefully because our body can come in contact with the current.
Thus, the precautions are to keep high voltage lines away from one socket. Use only a few high-power electric appliances at once.
To learn more about electric circuits, refer to the below link:
brainly.com/question/28221759
#SPJ4
This question is incomplete, the complete question is;
Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.
Use Fy=50 ksi and assume Cb=1.0 (if needed).
Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft
Explanation:
Given that;
section W 21 x 73 steel beam;
now from the steel table table for this section;
Zx = Sx = 151 in³
also given that; fy = 50 ksi and Cb = 1.0
QMn = 0.9 × Fy × Zx
so we substitute
QMn = 0.9 × 50 × 151
QMn = 6795 k-inch
we know that;
12inch equals 1 foot
so
QMn = 6795 k-inch / 12
QMn = 566.25 f-ft
Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft
Answer:
I = 8.3 Amp
potential drop = 83 V
Explanation:
Power = 100 KW
V = 12,000 V
R = 10 ohms
a)
Calculate current I in each wire:
P = I*V
I = P / V
I = 100 / 12 = 8.333 A
b)
Calculate potential drop in each wire:
V = I*R
V = (8.3) * (10)
V = 83 V