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I need help with the following homework problem.

Please Help

What I have done so far ......

Data

Voltage Supplied= 50V Load Voltage= 50/3 = 16.66V R1=? R2=? RL=? P(Load) = 1.0 mW VR1=? VR2=? IR1=? IR2=? IRL=?

Solution

Finding RL

P=V^2/RL RL= V^2/P RL= (16.66)^2/1.0 mW RL=277.5 kilo ohms

VL=VR2=16.66V

VR1=50V-16.66V=33.34V

IRL=VL/RL IRL=16.66/277.5 IRL=0.0600mA

IR1,IR2,R1 And R2 ?

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    \$\begingroup\$ What have you done so far? In which step or concept did you get stuck? \$\endgroup\$ – jDAQ Jun 18 '20 at 18:06
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    \$\begingroup\$ Edit your question and add all that to it. Did you find the value of \$R_L\$? And even if you didn't, show all the steps/solution attempts you have done. \$\endgroup\$ – jDAQ Jun 18 '20 at 18:11
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    \$\begingroup\$ I have edited the title to reflect the content of the question rather than "Help Please Help" which tells us nothing. \$\endgroup\$ – Transistor Jun 18 '20 at 18:14
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    \$\begingroup\$ Let me ask you some leading questions to help your train of thought. What’s the voltage across R2 and Rl? Consequently, what’s the voltage across R1? \$\endgroup\$ – winny Jun 18 '20 at 18:23
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    \$\begingroup\$ Voltage across R1 and R2 is not given :( but I guess Voltage across R2 and RL will be same as both these resistors are in parallel =16.66V \$\endgroup\$ – Prince Vegas Jun 18 '20 at 18:25
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You seem to have answered your question in the comment above. It's easy enough to calculate the value for \$R_L\$ given the voltage and power: $$R_L = V_L^2/P_L = 277.8 k\Omega$$ What may have been stumping you is that the problem has an extra degree of freedom; you can choose the value for either \$R_1\$ or \$R_2\$ and calculate the other. Your answer above is correct, but a simpler solution can be had if you choose \$R_2 = R_L\$, then \$R_2|| R_L =R_L/2\$, and since \$V_L\$ is \$1/3\$ of the supplied voltage, the required value of \$R_1\$ is twice that parallel combination, so \$R_1=R_L\$ as well.

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