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GATE 2015 EE - SET 1 - Question 41

Solution :

\frac{dC_{1}}{dP_{1}} = \frac{dC_{2}}{dP_{2}}

P_{1}+P{2}= 200MW -----------(1)

\frac{dC_{1}}{dP_{1}} = \frac{dC_{2}}{dP_{2}}

(0.01)(2)P_{1}+30 = (0.05)(2)P_{2}+10

0.02 P_{1} + 30 = 0.1 P_{2} + 10

0.1 P_{2} - 0.02 P_{1} = 20 --------(2)

Solving (1) and (2)

P_{1} =0 ,P_{2} =200

P_{1} = 100MW , P_{2} = 100 MW

\frac{dC_{2}}{dP_{2}} = 0.1 P_{2} + 10 = 0.1(100)+10 = 10 + 10 = 20

 

 

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