Your load voltage and arc voltage measurement should be:

3

liq [111]

Answer:

a

Explanation:

3 0

4 years ago

The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determi

kirza4 [7]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

$Y=\frac{1}{Z}$

Hence for each we have,

$Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S$

for the second impedance we have

$Y_{2}=\frac{1}{10}\\Y_{2}=0.1S$

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

$V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\$

$V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v$

The real power in the impedance is calculated as

$P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W$

for the second impedance

$P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w$

b. We determine the equivalent admittance

$Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\$

We convert the equivalent admittance back into the polar form

$Y_{total}=0.28\leq -19.65\\$

the source current flows is

$I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A$

6 0

3 years ago

What will the following segment of code output? score = 95; if (score > 95) cout << "Congratulations!\n"; cout <<

Anarel [89]

Answer:

That's a high score!

This is a test question!

Explanation:

The reason these two lines are printed and not the first one is simple. After the 'IF' condition has been stated, there is no use of parenthesis such as { and } to enclose the next lines. This means that only the first line after the 'IF' condition may be read or skipped depending on whether the condition (score>95) is met. Since the score is not larger than 95, and the 'IF' condition fails, the line 'Congratulations!' is not printed. The next two lines of the code are read as normal because they do not depend on the 'IF' condition.

5 0

3 years ago

Design an Armstrong indirect FM modulator to generate an FM signal with a carrier frequency 98.1 MHz and a frequency deviation △

Bess [88]

Answer:

See explaination

Explanation:

In the Armstrong method of FM generation, the phase of the carrier is directly modulated in the combing network through summation, generating indirect frequency modulation.

Very high frequency stability is achieved through Armstrong method since the crystal oscillator is used as carrier frequency generator.

Please kindly check attachment for the step by step solution of the given problem.

3 0

4 years ago

7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 v

Naddik [55]

Answer:

23.34 seconds

Explanation:

Flow rate = 1500

Arrival = 800 vehicle per hour

Cycle c = 60 seconds

Dissipation time = 10 seconds

Arrival time = 800/3600 = 0.2222

Rate of departure = 1500/3600 = 0.4167

Traffic density p = 0.2222/0.4167 = 0.5332

Real time = r

r + to + 10 = c

to = c-r-10 ----1

t0 = p*r/1-p ----2

Equate both 1 and 2

C-r-10 = p*r/1-p

60-r-10 = 0.5332r/1-0.5332

50-r = 0.5332r/0.4668

50-r = 1.1422r

50 = 1.1422r + r

50 = 2.1422r

r = 50/2.1422

r = 23.34 seconds

7 0

3 years ago