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UkoKoshka [18]
3 years ago
11

Using the two circuits, calculate i1, i2, v1, v2, the power (PR) being dissipated (absorbed) by the resistor, and the conductanc

e (PG). Assume that Is= 2.7A, Vs= 1.5 V, R= 270 Ω, and G= 0.27 S
Engineering
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

i_1=2.7\:\:A\\i_2=0.405\:\:A\\v_1=729\:\:V\\v_2=v_S=1.5\:\:V\\

Explanation:

i_1=i_S=2.7\:\:A\\v_1=i_1R=2.7*270=729\:\:V\\v_2=v_S=1.5\:\:V\\i_2=v_2G=1.5*0.27=0.405\:\:A

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Determine whether or not each of the following signals is periodic.
Sloan [31]

Answer:

a) periodic (N = 1)

b) not periodic

c) not periodic

d) periodic (N = 8)

e) periodic (N = 16)

Explanation:

For function to be a periodic: f(n) = f(n+N)

a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)

It is periodic with fundamental period N = 1

b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k

N must be integer. So it is nor periodic

c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16

Since N is dependent to n. So it is not periodic.

d) x[n]=cos(\frac{\pi }{2}  n) cos(\frac{\pi }{4}  n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8\\

So it is periodic with fundamental period N = 8.

e) x[n]=2cos(\frac{\pi }{4}  n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4

So it is periodic with N = 16.

3 0
3 years ago
Draw a 3-D physical structure of an NMOS transistor. Label four terminals: body, drain, gate, and source. And also label silicon
Vanyuwa [196]

Answer:

Answer is attached.

Explanation:

A NMOS is a n-channel MOSFET or Metal Oxide

Semiconductor Field Effect Transistor. This type

of transistor might be an enhancement or

depletion type nMOS transistor designed using

layers of Metal-oxide, Silicon-oxide and Silicon

fabricated on a substrate.

4 0
3 years ago
3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
4 years ago
2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu
n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

5 0
3 years ago
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Elan Coil [88]
Thx :) so much :))))))
4 0
3 years ago
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