Title: GR8677.064Post by: conantee on January 06, 2010, 11:54:56 AM
64. An alternating current electrical generator has a fixed internal impedance and is used to supply power to a passive load that has an impedance , where , and . For maximum power transfer between the generator and the load, should be equal to
(A) (B) (C) (D) (E) Title: Re: GR8677.064Post by: conantee on October 11, 2010, 11:51:03 AM
Don't be scared by complex numbers. Just do everything as if they are real numbers. Then, when we want to extract the information from the calculation, just consider real part.
Suppose the voltage from generator is . This voltage goes across internal impedance (let's call in ) and passive load () in series. The current in the circuit is now just . Power to the load is just Now, . To avoid large amount of calculation, let's use hand-waving argument. In order to keep the norm of the denominator small (so that the power is large), the imaginary part should go to zero. Otherwise, it will introduce some phase and larger denominator. This sounds like cheating since we haven't taken account of at numerator. However, the denominator has power of two, while the numerator has only one. So, it's OK. Hence, to make imaginary part of be zero, The correct answer is (C). Only 22 of 100 people answer correctly. |