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ฟิสิกส์โอลิมปิก วิทยาศาสตร์โอลิมปิก ข้อสอบแข่งขัน ข้อสอบชิงทุน => GRE - Physics => Topic started by: conantee on January 06, 2010, 11:54:56 AM

Title: GR8677.064
Post by: conantee on January 06, 2010, 11:54:56 AM
64. An alternating current electrical generator has a fixed internal impedance R_g + jX_g and is used to supply power to a passive load that has an impedance R_g + j X_l, where j = \sqrt{-1}, R_g \neq 0, and X_g \neq 0. For maximum power transfer between the generator and the load, X_l should be equal to
(A) 0
(B) X_g
(C) -X_g
(D) R_g
(E) -R_g

Title: Re: GR8677.064
Post 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 \tilde{V}. This voltage goes across internal impedance (let's call in Z_1) and passive load (Z_2) in series.

The current in the circuit is now just \tilde{I} = \frac{\tilde{V}}{Z_{tot}} = \frac{\tilde{V}}{Z_1+Z_2}

Power to the load is just \tilde{P} = \tilde{I}^2 Z_2 = \tilde{V} \frac{Z_2}{(Z_1+Z_2)^2}

Now, Z_1 + Z_2 = (R_g + jX_g) + (R_g + jX_l) = 2R_g + j(X_g + X_l). 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 Z_2 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 Z_1 + Z_2 be zero, X_l = - X_g

The correct answer is (C). Only 22 of 100 people answer correctly.