Print Page - GR8677.064

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) $-R_g$

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.

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