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xxTIMURxx [149]
3 years ago
7

what is the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation

Engineering
1 answer:
Elza [17]3 years ago
4 0

Based on the engineering analysis, the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF) between the output and the adjust terminal."

The LM1117 generally has a 1.2V at 800mA of load current.

The LM1117 is known to be a progression of low dropout linear voltage regulators. It is adjustable and be assigned between the output voltage of 1.25 to 13.8 V using just two external resistors.

Hence, in this case, it is concluded that the correct answer is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF), between the output and the adjust terminal.

Learn more here: brainly.com/question/25646477

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Answer:

A. S0 = 1, S1 = 0, S2 = 0

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A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
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Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

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