The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current.

"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

[tex]Y=\frac{1}{Z}[/tex]

Hence for each we have,

[tex]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[/tex]

for the second impedance we have

[tex]Y_{2}=\frac{1}{10}\\Y_{2}=0.1S[/tex]

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

[tex]V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\[/tex]

[tex]V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v[/tex]

The real power in the impedance is calculated as

[tex]P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W[/tex]

for the second impedance

[tex]P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w[/tex]

b. We determine the equivalent admittance

[tex]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\\[/tex]

We convert the equivalent admittance back into the polar form

[tex]Y_{total}=0.28\leq -19.65\\[/tex]

the source current flows is

[tex]I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A[/tex]