Thevenin's Problem: Solve for Voltage, Current, Power

[merchant]

Homework Statement

Homework Equations

V=IR
P=I^2R
maximum power when load resistance = thevenin's resistance

The Attempt at a Solution

I have been able to work out the new load resistance but none of the other values correctly.

 

[Svein]

1. Remove R_L. Now the voltage at the junction between 20Ω and 30Ω is \frac{30\Omega}{(20+30)\Omega}\cdot 15V = 9V with a source impedance of 20\Omega \left\lvert \right\rvert 30\Omega =\frac{20\cdot 30}{20 + 30}\Omega...

 

[merchant]

1. Remove R_L. Now the voltage at the junction between 20Ω and 30Ω is \frac{30\Omega}{(20+30)\Omega}\cdot 15V = 9V with a source impedance of 20\Omega \left\lvert \right\rvert 30\Omega =\frac{20\cdot 30}{20 + 30}\Omega...

Thank you for your reply, also how do you calculate the other values

 

[Svein]

Thank you for your reply, also how do you calculate the other values

I have given you the first steps. Now try the next: With R_L still disconnected, find the voltage and the source impedance at point A.