BOOK DETAILS
TITLE : "ELECTROMAGNETICS WITH APPLICATIONS"
EDITION : FIFTH EDITION(INDIAN EDITION)
AUTHORS : JOHN KRAUS, DANIEL FLEISCH
PRINT : THIRTEENTH REPRINT 2018
ISBN-13 : 978-0-07-070240-0
ISBN-13 : 0-07-070240-3
PUBLIC. : MC GRAW HILL EDUCATION
Considering page 141, viz. under chapter 3, "Transmission Lines", sub heading "The Terminated uniform transmission line and VSWR".
PROBLEM 3-4-2. Power Splitter. Referring to Fig. 3-15, two loads R1=300 Ω and R2 = 200 Ω are to be fed in phase from a 100 Ω line with R1 receiving twice the power of R2.
Find (a)l1 and Z1 and (b) l2 and Z2.
Here is my attempt,
$$ \frac{1}{100\angle0+z_1\angle{a_1}+300\angle0}=\frac{2}{100\angle0+z_2\angle{a_2}+200\angle0} $$
$$ {{100\angle0+z_2\angle{a_2}+200\angle0}}={{200\angle0+2z_1\angle{a_1}+600\angle0}} $$
$$ {\sqrt{{(300+z_2\cos{a_2})^2+(z_2\sin{a_2})^2}}}\angle{\arctan{\frac{z_2\sin{a_2}}{300+z_2\cos{a_2}}}}={\sqrt{{(800+2z_1\cos{a_1})^2+(2z_1\sin{a_1})^2}}}\angle{\arctan{\frac{2z_1\sin{a_1}}{800+2z_1\cos{a_1}}}} $$
$$ \text{On solving the angle and power part seperately} $$
$$ {\frac{z_2\sin{a_2}}{300+z_2\cos{a_2}}}={\frac{2z_1\sin{a_1}}{800+2z_1\cos{a_1}}} $$
$$ {{2z_1z_2\cos{a_1}\sin{a_2}}+{800z_2sina_2}}={{2z_1z_2\sin{a_1}\cos{a_2}}+{600z_1\sin{a_1}}} $$
$$ {2z_1z_2\sin{a_2-a_1}}={{600z_1\sin{a_1}}-{800z_2\sin{a_2}}} $$
or,
$$ {2z_1z_2\sin{a_1-a_2}}={{800z_2\sin{a_2}}-{600z_1\sin{a_1}}} $$
$$ \text{Now solving for magnitude,} $$
$$ {{(300+z_2\cos{a_2})^2+(z_2\sin{a_2})^2}}={{(800+2z_1\cos{a_1})^2+(2z_1\sin{a_1})^2}} $$
$${z_2^2+600z_2\cos{a_2}}={5500+3200z_1\cos{a_1}+4z_1^2} $$
Another attempt but different method,
$$\frac{1}{{100e^{j\cdot0}}+{z_1e^{j\cdot{a_1}}}+{300e^{j\cdot0}}}=\frac{2}{{100e^{j\cdot0}}+{z_2e^{j\cdot{a_2}}}+{200e^{j\cdot0}}}$$
$${{300e^{j\cdot0}}+{z_2e^{j\cdot{a_2}}}}={{800e^{j\cdot0}}+{2z_1e^{j\cdot{a_1}}}}$$
$${{z_2e^{j\cdot{a_2}}}-{2z_1e^{j\cdot{a_1}}}}={500e^{j\cdot0}}$$
After this I'm not trying. Either I don't know or too lazy. Posting answer is not worth if we don't know the solution.