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 ?
Best Answer
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.