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Max+power+theorem
- 1. BEF 12403 MaximumPower Transfer Theorem
- 2. Learning Outcomes: Upon completionof this unit the student will be able to: (i) Determine the conditions for maximum power transfer to any circuit element.
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- 4. Maximum Power TransferTheorem The power delivered by a voltage source or a current source is a function of its internal resistance and also of the load resistance. Maximum power is delivered by a source to its load when the load resistance is equal to the internal resistance of the source, that is, when Rin = RL E i Rint v RL Figure 2 Practical voltage source under test
- 5. Proof Consider Figure 2which shows a practical voltage source delivering power to a load resistor RL. The power delivered to RL is given by E i= E i Rint + RL vL RL Rint and that the load current is given by the expression Practical PL = i RL2 voltage source under test By combining the two equations above, we Figure 2 obtain the expression E2 PL = RL ( RL + Rint ) 2
- 6. Proof (continued) To findthe value of RL that maximises the expression for PL, we first need to differentiate the expression for PL with respect to RL and then equate it to zero. Computing the derivative, we obtain the following expression: E 2 ( RL + Rint ) − 2 E 2 RL ( RL + Rint ) 2 dPL = =0 E i dRL ( RL + Rint ) 4 v RL Rint which leads to the expression ( RL + Rint ) 2 − 2 E 2 RL ( RL + Rint ) = 0 Practical voltage source It is easy to verify that the solution of this under test equation is Figure 2 RL= Rint Thus, to transfer maximum power to a load, the equivalent source and load resistances must be matched, that is, equal to each other.
- 7. • This is called a "matched condition" and as a general rule, maximum power is transferred from an active device such as a power supply or battery to an external device when the impedance of the external device exactly matches the impedance of the source. • One good example of impedance matching is between an audio amplifier and a loudspeaker. • Improper impedance matching can lead to excessive power loss and heat dissipation.
- 8. Proof (continued) Figure 3depicts a plot of the load power PL divided by E2 versus the ratio of RL to Rint. Note that this value is maximum when RL = Rint. PL/E2 RL/Rint 0 RL = Rint Figure 3
- 9. When the conditionsof the maximum power transfer theorem are met, the total power delivered to the load is: PL = ( E / 2) 2 = E2 E i vL RL Rint 4 Rint Rint This is one half of the total power generated Practical by the circuit, half the power is absorbed in voltage source the resistance RINT. under test Figure 2
- 10. Worked Example Given thecircuit in Figure 4, where: RS = 25Ω RL is variable between 0 - 100Ω VS = 100v Complete the following table to determine the current and power in the circuit for different values of load resistance. Figure 4
- 11. Solution Graph of Poweragainst Load Resistance:
- 12. Solution (continued) From theabove table and graph we can see that the Maximum Power Transfer occurs in the load when the load resistance, RL is equal in value to the source resistance, RS that is: RS = RL = 25 Ω.
- 13. Exercise Plot the powerdissipation of the load resistance, for several values between 1 kΩ and 20 kΩ: At what load resistance value is the load's power dissipation maximized? Figure 4
- 14. Answer Figure 5
- 15. Worked Example Suppose wewere planning to use a photovoltaic panel to generate electricity and electrolyze water into hydrogen and oxygen gas: Our goal is to electrolyze as much water as possible, and this means we must maximize the electrolysis cell's power dissipation. Explain how we could experimentally determine the optimum internal resistance of the electrolysis cell, prior to actually building it, using nothing but the solar panel, a rheostat, and a DMM (digital multimeter).
- 16. Answer Experimentally determine whatamount load resistance drops exactly one-half of the panel's open-circuit voltage.
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