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nordsb [41]
2 years ago
5

What is the maximum powr of a module in Watts to the nearest whole Watt?​

Physics
1 answer:
KonstantinChe [14]2 years ago
6 0

Complete question is;

You are looking at a module specification

sheet that has the table of information

below. What is the maximum power of this

module in Watts to the nearest whole Watt?

Value

Polycrystalline si

Characteristic

Cell Type

Cell

Configuration

Voc

160 in series

137.2 V

V_imp: 29.3 V

Ilsc: 8.60 A

I_Imp: 8.02 A

Dimensions (mm/in): 1000 x 1600 x 50 mm / 39.4" x 63" x 2"

Weight: 10 kg / 22 lbs​

Answer:

P ≈ 235 Watts

Explanation:

Formula for power is;

P = IV

Now, for maximum power, we will make use of I_imp and V_imp given

Thus, P = I_imp × V_imp

We are given;

V_imp: 29.3 V

I_Imp: 8.02 A

Thus: P = 8.02 × 29.3 = 234.986 Watts

We are to approximate to the nearest whole watt.

Thus: P ≈ 235 Watts

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<h2>Answer:</h2>

\boxed{P=96.09MW}

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega

So we can calculate the power loss as:

P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}

Finally, the power loss due to resistance in the line is 96.09MW

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