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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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Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500
Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

return 0;

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Explanation:

7 0
3 years ago
1. Which of the following is NOT an example of a lever?
Umnica [9.8K]
A staircase what makes it a lever is another objects used to displace the force better
8 0
3 years ago
The volume fraction of particles in a WC-particle Cu-matrix CERMET is 0.84 Calculate the minimum expected elastic modulus, in GP
Lynna [10]

Answer:

the minimum expected elastic modulus is 372.27 Gpa

Explanation:

First we put down the data in the given question;

Volume fraction V_f = 0.84

Volume fraction of matrix material V_m = 1 - 0.84 = 0.16

Elastic module of particle E_f = 682 GPa

Elastic module of matrix material E_m = 110 GPa

Now, the minimum expected elastic modulus will be;

E_{CT = (E_f × E_m ) / ( E_fV_m  + E_m V_f  )

so we substitute in our values

E_{CT = (682 × 110 ) / ( [ 682 × 0.16 ]  + [ 110 × 0.84] )

E_{CT = ( 75,020 ) / ( 109.12 + 92.4 )

E_{CT = 75,020 / 201.52

E_{CT = 372.27 Gpa

Therefore, the minimum expected elastic modulus is 372.27 Gpa

7 0
3 years ago
Using six bits, Show the binary counting sequence from 000000 to 111111
fenix001 [56]

Answer:

Explanation:

The least significant unit bits switch from 0 → 1 or 1 → 0 with each count.

The next bit(two-position) which is the second bit usually be at 0s for two counts and at  1s at two counts. The third bit (four positions) usually be at 0s for four counts and then at 1 for four counts. etc. The fourth bit (eight positions) usually be at 0s for eight counts and then at 1 for eight counts and so on. The fifth bit (sixteen positions) usually be at 0s for sixteen counts and then at 1 for sixteen counts and so on. The sixth bit (thirty-two positions) usually be at 0s for thirty-two counts and then at 1 for thirty-two counts and so on in the sequence of six-bit binary counting.

From the document attached below, we can see a document file showing the binary counts sequences from 000000 to 111111.

Download docx
8 0
3 years ago
Determine the magnitude of force P needed to start towing the 40kg crate.Also determine the location of the resultant normal for
erastova [34]

The distance from Point A=500 mm

Explanation:

M := 40kg c := 200mm

μs := 0.3 d := 3

a := 400mm e := 4

b := 800mm

Initial guesses: N_{c} := 200 N P := 50N

Plz refer to the image below

5 0
3 years ago
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