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III. Long Answer Questions · Q19

Q.Give the advantage of AC in long distance power transmission with an illustration.

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Step 1. Transmitting 2 MW at V=10V=10 kV over lines of resistance R=40 ΩR=40\ \Omega: current I=P/V=(2×106)/(10×103)=200I=P/V=(2\times10^6)/(10\times10^3)=200 A.

Step 2. Power loss =I2R=(200)2(40)=1.6×106=I^2R=(200)^2(40)=1.6\times10^6 W =1.6=1.6 MW, which is (1.6/2)×100%=80%(1.6/2)\times100\%=80\% of the transmitted power -- an unacceptably large loss.

Step 3. Transmitting the same 2 MW instead at V=100V=100 kV: current I=(2×106)/(100×103)=20I=(2\times10^6)/(100\times10^3)=20 A.

Step 4. Power loss =I2R=(20)2(40)=0.016×106=I^2R=(20)^2(40)=0.016\times10^6 W, which is (0.016/2)×100%=0.8%(0.016/2)\times100\%=0.8\% of the transmitted power. …

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