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

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Homework Statement



vuZohAF.png

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

Thank you for your reply, also how do you calculate the other values
 
merchant said:
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 RL still disconnected, find the voltage and the source impedance at point A.