Thevenins problem

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1. May 2, 2015

merchant

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution
I have been able to work out the new load resistance but none of the other values correctly.

2. May 2, 2015

Svein

1. Remove RL. 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 $20\Omega \left\lvert \right\rvert 30\Omega =\frac{20\cdot 30}{20 + 30}\Omega$...

3. May 2, 2015

merchant

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

4. May 2, 2015

Svein

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.