My Answer is as follows:
However, the correct answer is as shown below:
My question is where is my mistake?
So, I added the below as of my understanding is my reasoning accurate or not?
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1\$\begingroup\$ You forgot to consider the angle when adding your impedances. \$\endgroup\$– HearthCommented Jan 6, 2021 at 5:40
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1 Answer
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Your correctly found that the real power of the total circuit is the sum of the real powers of the components. However, you made a mistake by assuming you could just add the the apparent powers (Volt-amps).
$$S_t \ne S_1 + S_2$$
(at least not if S is a scalar (which it usually is, and definitely is in this case) rather than a vector)
You need to calculate the reactive powers of each component and add them. The reactive power of the resistance is of course 0. The reactive power of the unknown load is calculated in the answer. You need to find that, and you didn't. Once you know the total reactive power, and the total real power, you can find the total apparent power and/or the power factor.
answered Jan 6, 2021 at 5:45
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\$\begingroup\$ Thank you. So if I got you right, what you're saying is the value 10 KVA for the pure resistive load is accurate. And the 40/0.8 or 50 KVA is also accurate. BUT, they don't add algebraically and thus we MUST use trigonometry properties to add them in order to find the TOTAL/COMBINED apparent power. \$\endgroup\$– OMARCommented Jan 6, 2021 at 6:20
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