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www.GOALias.blogspot.com7.23 4007.24 Hydroelectric power = h ρ g × A × v = h ρ g βwhere β = Av is the flow (volume of water flowing per second across across-section).Electric power available = 0.6 × 300 × 10 3 × 9.8 × 100 W= 176 MW7.25 Line resistance = 30 × 0.5 = 15 Ω.800 × 1000 Wrms current in the line == 200 A4000 V(a) Line power loss = (200 A) 2 × 15 Ω = 600 kW.(b) Power supply by the plant = 800 kW + 600 kW = 1400 kW.(c) Voltage drop on the line = 200 A × 15 Ω = 3000 V.The step-up transformer at the plant is 440 V – 7000 V.800 × 1000 W7.26 Current = = 20 A40,000 V(a) Line power loss = (20 A) 2 × (15 Ω) = 6 kW.(b) Power supply by the plant = 800 kW + 6 kW = 806 kW.(c) Voltage drop on the line = 20 A × 15 Ω = 300 V.The step-up transformer is 440 V – 40, 300 V. It is clear thatpercentage power loss is greatly reduced by high voltagetransmission. In Exercise 7.25, this power loss is (600/1400)× 100 = 43%. In this exercise, it is only (6/806) × 100 = 0.74%.AnswersCHAPTER 88.1 (a) C = ε0 A/d = 80.1 pFdQdtdVdtdV= Cdt0.15=80.1 × 10–129 –1= 1.87 × 10 V sd(b) i d= ε0 Φ Ε .. Now across the capacitor ΦdtE= EA, ignoring endcorrections.dΦΕTherefore, id= ε0A d tQNow, E = . Therefore,ε A0dEdti= , which implies iε Ad= i = 0.15 A.0(c)Yes, provided by ‘current’ we mean the sum of conduction anddisplacement currents.8.2 (a) I rms= V rmsωC = 6.9µA305