Assume that a p+ - n diode is built with an n region width l smaller than a hole diffusion length (l

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

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

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