# Voltage divider homework

I need help with the following homework problem.

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 ?

• What have you done so far? In which step or concept did you get stuck? – jDAQ Jun 18 '20 at 18:06
• 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. – jDAQ Jun 18 '20 at 18:11
• I have edited the title to reflect the content of the question rather than "Help Please Help" which tells us nothing. – Transistor Jun 18 '20 at 18:14
• 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? – winny Jun 18 '20 at 18:23
• 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 – Prince Vegas Jun 18 '20 at 18:25

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.