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xaer04
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[SOLVED] Electric Current
"A power plant produces 1000 MW to supply a city 40km away. Current flows from the power plant on a single wire of resistance .050[itex]\Omega/km[/tex], through the city, and returns via the ground, assumed to have negligible resistance. At the power plant the voltage between the wire and ground is 115kV. a.) What is the current in the wire? b.) What fraction of the power is lost in transmission?"
Initial Power [itex]= P_0 = 1.0x10^9[/itex]
Wire length [itex]= L = 40km[/itex]
Resistance per unit length [itex]= R/L = .050 \Omega /km[/itex]
->Resistance for 40km wire [itex]= 2\Omega[/itex]
Voltage between wire and ground [itex] = 115x10^3V[/itex]
Electric Power
[tex]P = IV[/tex]
[tex]P = I^2 R[/tex]
[tex]P = \frac{V^2}{R}[/tex]
Electric Current
[tex]I=\frac{V}{R}[/tex]
I tried several of these and got all different answers.
[tex]I = P/V = \frac{1x10^9W}{1.15x10^5V} = 8695.7 A[/tex]
[tex]I = \sqrt{P/R} = \sqrt{(1x10^9W)/(2\Omega)} = 22360.7 A[/tex]
[tex]I = V/R = (1.15x10^5V)/(2\Omega) = 57500 A[/tex]
I couldn't even get the power to work out using the remaining Power equation, reliant on Volts and Resistance.
[tex]P = (V^2)/(R) = (1.15x10^5V)^2/(2\Omega) = 6.61x10^9W[/tex]
Please help me understand this:(
Homework Statement
"A power plant produces 1000 MW to supply a city 40km away. Current flows from the power plant on a single wire of resistance .050[itex]\Omega/km[/tex], through the city, and returns via the ground, assumed to have negligible resistance. At the power plant the voltage between the wire and ground is 115kV. a.) What is the current in the wire? b.) What fraction of the power is lost in transmission?"
Initial Power [itex]= P_0 = 1.0x10^9[/itex]
Wire length [itex]= L = 40km[/itex]
Resistance per unit length [itex]= R/L = .050 \Omega /km[/itex]
->Resistance for 40km wire [itex]= 2\Omega[/itex]
Voltage between wire and ground [itex] = 115x10^3V[/itex]
Homework Equations
Electric Power
[tex]P = IV[/tex]
[tex]P = I^2 R[/tex]
[tex]P = \frac{V^2}{R}[/tex]
Electric Current
[tex]I=\frac{V}{R}[/tex]
The Attempt at a Solution
I tried several of these and got all different answers.
[tex]I = P/V = \frac{1x10^9W}{1.15x10^5V} = 8695.7 A[/tex]
[tex]I = \sqrt{P/R} = \sqrt{(1x10^9W)/(2\Omega)} = 22360.7 A[/tex]
[tex]I = V/R = (1.15x10^5V)/(2\Omega) = 57500 A[/tex]
I couldn't even get the power to work out using the remaining Power equation, reliant on Volts and Resistance.
[tex]P = (V^2)/(R) = (1.15x10^5V)^2/(2\Omega) = 6.61x10^9W[/tex]
Please help me understand this:(