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Let us examine a simple model of an electric power system. A power plant provides energy using a generator with an output voltage VA 141 cos(400t) kV. The inductance L represents the effect = of all power lines and transformers bringing the power from the plant to a customer. The customer's load is represented by a resistance RL and a capacitance C. With no capacitor (i.e. the branch with the capacitor in the diagram below is open and the customer only has a resistive load RL = 100 ): 1. Determine VB using the impedance method (an expression and an actual # for VB). 2. Determine the average power dissipated by the resistive load. Recall that for sinusoidal sources, P = |VB |²/2RL- In an attempt to improve the power delivered to the load resistor RL, the customer adds a capacitor C in parallel with RL as shown in the diagram below. The customer finds that C = 25 μF works well. 3. Determine VB using the impedance method (an expression and an actual # for VB). 4. Determine the average power dissipative in RL (RL = 100 Q still). 5. Should the customer be happy with this new configuration in terms of power delivery? 6. At the end of the day, the customer turns off 90% of the load to save energy (making RL = 1 ks), at which point the remaining equipment that is still connected and turned on begins to spark and catch fire. Why? VA L 0000 RL www + VB Power plant Power line Customer load

Let us examine a simple model of an electric power system. A power plant provides energy using a generator with an output voltage VA 141 cos(400t) kV. The inductance L represents the effect = of all power lines and transformers bringing the power from the plant to a customer. The customer's load is represented by a resistance RL and a capacitance C. With no capacitor (i.e. the branch with the capacitor in the diagram below is open and the customer only has a resistive load RL = 100 ): 1. Determine VB using the impedance method (an expression and an actual # for VB). 2. Determine the average power dissipated by the resistive load. Recall that for sinusoidal sources, P = |VB |²/2RL- In an attempt to improve the power delivered to the load resistor RL, the customer adds a capacitor C in parallel with RL as shown in the diagram below. The customer finds that C = 25 μF works well. 3. Determine VB using the impedance method (an expression and an actual # for VB). 4. Determine the average power dissipative in RL (RL = 100 Q still). 5. Should the customer be happy with this new configuration in terms of power delivery? 6. At the end of the day, the customer turns off 90% of the load to save energy (making RL = 1 ks), at which point the remaining equipment that is still connected and turned on begins to spark and catch fire. Why? VA L 0000 RL www + VB Power plant Power line Customer load

Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.27P: An industrial load consisting of a bank of induction motors consumes 50 kW at a power factor of 0.8...
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Let us examine a simple model of an electric power system. A power plant provides energy using a generator with an output voltage VA 141 cos(400t) kV. The inductance L represents the effect = of all power lines and transformers bringing the power from the plant to a customer. The customer's load is represented by a resistance RL and a capacitance C. With no capacitor (i.e. the branch with the capacitor in the diagram below is open and the customer only has a resistive load RL = 100 ): 1. Determine VB using the impedance method (an expression and an actual # for VB). 2. Determine the average power dissipated by the resistive load. Recall that for sinusoidal sources, P = |VB |²/2RL- In an attempt to improve the power delivered to the load resistor RL, the customer adds a capacitor C in parallel with RL as shown in the diagram below. The customer finds that C = 25 μF works well. 3. Determine VB using the impedance method (an expression and an actual # for VB). 4. Determine the average power dissipative in RL (RL = 100 Q still). 5. Should the customer be happy with this new configuration in terms of power delivery? 6. At the end of the day, the customer turns off 90% of the load to save energy (making RL = 1 ks), at which point the remaining equipment that is still connected and turned on begins to spark and catch fire. Why? VA L 0000 RL www + VB Power plant Power line Customer load
Transcribed Image Text:Let us examine a simple model of an electric power system. A power plant provides energy using a generator with an output voltage VA 141 cos(400t) kV. The inductance L represents the effect = of all power lines and transformers bringing the power from the plant to a customer. The customer's load is represented by a resistance RL and a capacitance C. With no capacitor (i.e. the branch with the capacitor in the diagram below is open and the customer only has a resistive load RL = 100 ): 1. Determine VB using the impedance method (an expression and an actual # for VB). 2. Determine the average power dissipated by the resistive load. Recall that for sinusoidal sources, P = |VB |²/2RL- In an attempt to improve the power delivered to the load resistor RL, the customer adds a capacitor C in parallel with RL as shown in the diagram below. The customer finds that C = 25 μF works well. 3. Determine VB using the impedance method (an expression and an actual # for VB). 4. Determine the average power dissipative in RL (RL = 100 Q still). 5. Should the customer be happy with this new configuration in terms of power delivery? 6. At the end of the day, the customer turns off 90% of the load to save energy (making RL = 1 ks), at which point the remaining equipment that is still connected and turned on begins to spark and catch fire. Why? VA L 0000 RL www + VB Power plant Power line Customer load
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