# 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.

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