Where should the Q point of an amplifier be on a DC load line to have the LARGEST linear output?

Answers

Answer 1
Answer:

Answer:

At the middle of the DC load

Explanation:

For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region


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Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string is your own creative choice. It can be names, favorite foods, animals, anything. Just make it up yourself. Do not copy the string from another source. Turn the string into a list of words using split. Delete three words from the list, but delete each one using a different kind of Python operation. Sort the list. Add new words to the list (three or more) using three different kinds of Python operation. Turn the list of words back into a single string using join. Print the string.

Answers

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

\n Then I used methods append(), insert() and extend() for adding elements to the list.

\n Finally I converted list into a string using join() and adding space in between the elements of the list.

What is the function of deaerator in thermal power plant? ​

Answers

Answer:

The function of the Deaerator is to remove dissolved non-condensable gases and to heat boiler feed water.

Explanation:

Select the true statements regarding rigid bars. a. A rigid bar can bend but does not change length.
b. A rigid bar does not bend regardless of the loads acting upon it.
c. A rigid bar deforms when experiencing applied loads.
d. A rigid bar is unable to translate or rotate about a support.
e. A rigid bar represents an object that does not experience deformation of any kind.

Answers

Answer:

option b and E are true

Explanation:

A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:

option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.

option b is true. A rigid bar remains unchanged regardless of the load that it carries.

option c is incorrect, a rigid bar does not deform with loads on it

option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.

option e is true. A rigid bar would not experience any deformation whatsoever.

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.

Answers

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages

Answers

Answer:

Explanation:

A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.

When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.

Given that:

input (primary) voltage of the transformer, V_i=220~V

no. of turns in the primary coil, N_i=230

  • When the output voltage is 5.60 V:

V_o=5.60~V

(N_i)/(N_o) =(V_i)/(V_o)

(N_o)/(230)=(5.60)/(220)

N_o=5.85\approx 6 turns compensating the losses

  • When the output voltage is 12.0 V:

V_o=12.0~V

(N_i)/(N_o) =(V_i)/(V_o)

(N_o)/(230)=(12.0)/(220)

N_o=12.45\approx 13 turns compensating the losses

  • When the output voltage is 480 V:

V_o=480~V

(N_i)/(N_o) =(V_i)/(V_o)

(N_o)/(230)=(480)/(220)

N_o=501.8\approx 502 turns compensating the losses