S13Phys2BaLec26B

# Lets make some estimations r 10 ω v 100000v to

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Let’s make some estimations: R = 10 Ω , V = 100,000V to deliver 100kW of power. First, don’t make a common mistake: This equation can only be used with a potential difference ( Δ V) across a resistor. <-no!

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Power Instead we will find the current by the power transfer equation. Now we can input this current in the proper power dissipation equations. This only represents a 0.01% loss of power transferred to resistive effects (not bad!). P = IV
Power What if power companies used lower potential for their transmission lines (say 2,000V)? Now we can input this current in the proper power dissipation equations. 25% of power is lost to heat. That isn’t good business. P = IV

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Power dissipation C J P = I V s C [W] ( V = IR ) P = I 2 R ( V ) 2 P = R V I = R ( )
Superconductors Supercondutors can help solve these problems. Superconductors are materials whose resistance falls to virtually zero below a certain critical temperature, T C . Once a current is set up in a superconductor, it persists without any applied potential difference.

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Superconductors Above T c , the superconductor acts as a normal metal. Unfortunately superconductors currently only exist at low temperatures (highest ~150K) [–123 o C]. But we are getting closer to room temperature superconductivity. In the mid-80’s the highest was near 30K [–243 o C].
Circuits Light bulbs are resistors that you can measure the power dissipated by them with their brightness.

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