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,\$
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:
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?