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Mice21 [21]
3 years ago
8

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

uctor, or a capacitor. At time t = 0 the voltage is zero and increasing toward a maximum. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -I max, where I max is the current amplitude. What is the unknown element?
Engineering
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

The unknown element is Capacitor.

Explanation:

The sinusoidal voltage is given as:

v(t) = V Sin (ωt + Ф)

Where:

V = Amplitude of Voltage

ω = 2π / T = > Time period (T)

Ф = phase shift

Considering no horizontal phase shift in the wave form, the equation can be written as:

v(t) = V Sin (ωt)--------(1)

Since, current in the capacitor can be given as:

i(t) = C dv(t)/dt = ωCV Cos (ωt)--------(2)

Now, checking all conditions:

At t=0 :

Equation (1) implies:

v(t) = V Sin [(2π/T)(0)] = V Sin (0)

v(t) = 0

The above finding satisfies the condition in the question. Now checking other conditions.

At t = T/4:

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/4)] = ωCV Cos [(π/2)] = ωCV (0)

i(t) = 0

At t = T/2 :

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/2)] = ωCV Cos [(π)] = ωCV (-1)

i(t) = - ωCV = max amplitude of current in negative direction

All three conditions of voltage and currents of question are satisfied with equations of capacitor hence, the unknown element is capacitor.

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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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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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