Why will i calculate the wrong power ,but right loading resistance in this circuit?

Question

The question wants me to calculate that what is the \$R_L\$ value when we have the maximum \$P_{R_L}\$ from this circuit

^{simulate this circuit – Schematic created using CircuitLab}

My thinking

When we are calculating the equivalent resistance,we can see the original circuit as the circuit in the left hand below ,and the left hand circuit is equal to the right hand circuit,by the way,\$R_{10}=R_1//R_3,\$ and \$R_{12}=R_2//R_4,\$

^{simulate this circuit}

So according to \$P_{R_L}=I^2 R_L=(\frac{60}{2.1+R_L+0.9})^2R_L=\frac{60^2}{(3+R_L)^2}R_L\$,and use the Maximum Power Transfer Theorem,we can know \$R_L=3Ω\$, we can have the maximum \$P_{R_L}\$ from this circuit.

So now we can know the voltage of \$a\$ point,\$V_a\$, and the voltage of \$b\$ point,\$V_b\$ is as below

\$V_a=60\frac{3+0.9}{2.1+3+0.9}=39V\$

\$V_b=60\frac{0.9}{2.1+3+0.9}=9V\$

So the voltage of the loading resistor is \$39-9=30\$,so the \$P_{R_L}=\frac{V^2}{R_L}=\frac{30^2}{3}=300W \$

However,the answer shows me

Answer:

^{simulate this circuit}

The \$V_{th}=60\frac{3}{7+3}-60\frac{1}{9+1}=12\$,and \$R_{th}=(7//3)+(9//1)=3 \$ ,so \$P_{R_L}=\frac{V^2}{4R_{th}}=12\$

I found that the main difference between solution and my thinking is that our \$V_{th}\$ are not the same ,my \$V_{th}\$ is the voltage of the loading resistor=30V,but solution's is \$12V\$ ,Why can't the voltage of loading resistor be calculated as i think??where am i wrong?

Answer 1

You correctly found the Thevenin resistance of 3 ohm but then it went wrong when you assumed all the resistors were in series with 60 volts. The 3 ohms calculated is also the load you need to apply between A and B to get maximum power transfer too.

Why can't the voltage of loading resistor be calculated as i think??where am i wrong?

Consider that this can be solved by "splitting the voltage source" covered in E5 is this short document: -

And that circuit becomes the one below when you convert the potential dividers on each supply and, calculate the new voltage source value for each side: -

So, the difference voltage between the two sources is 12 volts and it is this 12 volts that drives current through the 3 ohm resistor connected across A and B. So, that's: -

$$I = \dfrac{12\text{ volts}}{2.1 + 3 + 0.9} = 2 \text{ amps}$$

So the maximum power into the Thevenin resistance of 3 ohms is 12 watts.