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:
As there was no attached picture, I will explain how to take the measurement of liquids in any buret which you can then apply to the specific question
Explanation:
A buret is a laboratory apparatus used to precisely measure the volume of liquids (usually alkalise or bases) used in a titration experiment. The standard buret has a capacity of 50 ml and graduated in 0.1ml though burets with smaller capacities exist.
From the question, your buret is filled to the top (0.00ml) with liquid. It is very important when taking buret readings to place the buret below your eye level so that the bottom meniscus (lower part of the liquid) can be read.
To take the buret reading, note your initial buret reading (in this case 0.00ml) then titrate the liquid base in the buret against the acid by opening the tap located at the bottom of the buret.
When the titration or reaction is complete, note the final reading against the calibration of buret. You can do this by observing the lower meniscus of the liquid remaining in the buret. (Remember to keep the buret at eye level to avoid parallax error),
The difference between your final buret reading and the initial buret reading gives you the precise volume of liquid used in the reaction.
Answer:
a) 
b) 
c)
Explanation:
From the question we are told that:
Beaker Mass 
Liquid Mass 
Balance D:
Mass 
Balance E:
Mass 
Volume 
a)
Generally the equation for Liquid's density is mathematically given by
b)
Generally the equation for D's Reading at A pulled is mathematically given by
m_d = mass of block - mass of liquid displaced
c)
Generally the equation for E's Reading at A pulled is mathematically given by
Answer:
Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing
Explanation:
The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item
Answer:
Z = 29.938Ω ∠22.04°
I = 2.494A
Explanation:
Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:
Z² = R²+(Xl-Xc)²
Z = √R²+(Xl-Xc)²
R is the resistance = 4Ω
Xl is the inductive reactance = ωL
Xc is the capacitive reactance =
1/ωc
Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Xl = 2000×6×10^-3
Xl = 12Ω
Xc = 1/2000×12×10^-6
Xc = 1/24000×10^-6
Xc = 1/0.024
Xc = 41.67Ω
Z = √4²+(12-41.67)²
Z = √16+880.31
Z = √896.31
Z = 29.938Ω (to 3dp)
θ = tan^-1(Xl-Xc)/R
θ = tan^-1(12-41.67)/12
θ = tan^-1(-29.67)/12
θ = tan^-1 -2.47
θ = -67.96°
θ = 90-67.96
θ = 22.04° (to 2dp)
To determine the current, we will use the relationship
V = IZ
I =V/Z
Given V = 12V
I = 29.93/12
I = 2.494A (3dp)
