Recent content by Tautomer22
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P = VI = (281V)(3.75A) = 1054 W
So you're saying I need to focus on the turns ratio of the transformer rather than the resistance in ohms? I understand what you're saying, but I suppose I attacked this problem under the assumtpion that I would solve it by 1) Finding the power at the power plant 2) Finding the power...- Tautomer22
- Post #5
- Forum: Introductory Physics Homework Help
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P = VI = (281V)(3.75A) = 1054 W
I understand your logic, but it's the numbers I don't understand... With an resistance of 75 ohms and a current of 3.75, the voltage I calculate is = 281 V (which is very far off from the expected 240,000 V). Also, if I simply use a power equation, such as I2R then I get a power equal to 1054 W...- Tautomer22
- Post #3
- Forum: Introductory Physics Homework Help
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P = VI = (281V)(3.75A) = 1054 W
Homework Statement A small power plant produces a voltage of 6.0 kV and 150 A. The voltage is stepped up to 240 kV by the transformer before it is transmitted to a substation. The resistance of the transmission line between the power plants and the substation is 75 Ohms. What percentage...- Tautomer22
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- Lines Loss Power Power lines Power loss
- Replies: 6
- Forum: Introductory Physics Homework Help
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Electic Flux through a cube Why am I getting a value?
I was unclear on this point, thank you so much gneil! That solves the whole issue for me. I knew I was missing a basic idea here, and I wasn't deducing it from solutions to similar problems (alas).- Tautomer22
- Post #3
- Forum: Introductory Physics Homework Help
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Electic Flux through a cube Why am I getting a value?
Homework Statement A cubical Gaussian surface is placed in a uniform electric field as shown in the figure to the right. The length of each edge of the cube is 1.0 m. The uniform electric field has a magnitude of 5.0 * 10^8 N/C and passes through the left and right sides of the cube...- Tautomer22
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- Cube Flux Value
- Replies: 2
- Forum: Introductory Physics Homework Help