A copper wire with a criss section of 5 mm^2 and 2 km long has an effective resistance of 2Ohms.
a) Find the power lost in this wire if 10000 W of power is put through it with 120 V of potential. [hint: use P=VI to get current first then P=I^2R]
b) Express your answer from a) as a percentage of the total power.
c) Find the power lost if 10000 W is put through at 1200 V insteaf.
d) Express your answer from c) as a percentage of the total.
e) High-voltage transmission lines reach 155 kV of potential. based on your answers to a) - d), what can you say about this type of power line?
I thought I knew how to do this, but for a) I got 13888.89, which doesn't make sense since that's bogger than the original P I was given. I'm not asking for all the questions to be nswered, but I'd really appreciate just an explanation and correction so as to how I got 13888.89 for a). Thank you in advance. :)
The resisitivity of copper is 1.7*10^-6 ohm-cm, so the resistance of that wire should be (1.7*10^-6 ohm cm)*(2*10^5 cm)*(0.5 cm^2) = 0.17 ohms. That is not consistent with the 2 ohms that they claim is the resistance. Perhaps they meant 0.2 ohms
The current is I = P/V = 10,000/120 = 83.33 A and the dissipated power in wires is I^2 R = 13,889 W.
Something is wrong with the problem statement. I don't understand why they would give you the wire dimensions amd material, and then tell you an inconsistently high value for the resistance.
The term "effective resistance" is sometimes used for P/I^2 in AC lines with high frequencies and large diameters, when magnetic effects cause the current to be nonuniformly distributed in the wire. That is probably not the case here.
oh, ok, I see.
Well, at least I see that the answer I got of 13889 was right. Okay, thank you very much for your explanaiton. :)
I agree the problem probably meant for resistance to be .2 ohms, it was probably a typographical error in the text.
One note on effective resistance. In power transmission or distribution lines, effective resistance includes copper losses due to resistance, but also includes energy losses do to magnetic losses in attached transformers, either hysteresis or eddy currents. Typically this almost doubles resistance losses,giving a real power loss on the line of about twice what the ohmic resistance would be, but certainly depends on the number of transformers attached to the lines. In distribution circuits, these losses are noteworthy.
To find the power lost in the copper wire, we need to determine the current passing through it first. Let's break down the steps for question a):
Given:
- Potential difference (V) = 120 V
- Power (P) = 10000 W
- Resistance (R) = 2 Ω
- Cross-sectional area (A) = 5 mm^2
Step 1: Calculate the current (I) using the formula P = VI.
P = VI
10000 W = 120 V * I
I = 10000 W / 120 V
I ≈ 83.33 A
Step 2: Calculate the power lost (P_loss) using the formula P_loss = I^2 * R.
P_loss = (83.33 A)^2 * 2 Ω
P_loss ≈ 13888.89 W (which is the value you obtained)
It seems that you made an error in the calculation of the current. The current should be approximately 83.33 A, not 13888.89 A.
To correct the mistake, let's recalculate the power lost in the wire:
P_loss = (83.33 A)^2 * 2 Ω
P_loss ≈ 138.89 W
Therefore, the power lost in the wire when 10000 W of power is put through it is approximately 138.89 W, not 13888.89 W.
I hope this clarification helps! Let me know if you have any further questions.