Maximum Power dissipated at load resistor

[Bling Fizikst]

Homework Statement

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Relevant Equations

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So , max power is dissipated when the load resistance and thevenin resistance becomes equal such that $$P\leq \frac{V_{\text{th}}^2}{4R_{\text{th}}}$$
Then i applied the 'star delta' transformation by converting the 'delta' ABC to 'star' ABC .

From there , by loop current method , i got $$i=\frac{\epsilon}{\frac{5R}{3}+r}$$ (assuming the other $\frac{R}{3}$ is redundant)
Hence, $$V_{\text{th}}=i\cdot \frac{R}{3}=\frac{\epsilon R}{5R+3r}$$
Now , by removing the battery $\epsilon$ , i got $R_{\text{th}}=\frac{5R}{3}+r$

Computing , $P_{\max}$ using these values gives the wrong answer . Not sure where i went wrong . I have confusions regarding the value of $V_{\text{th}}$ though

 

[DaveE]

Is $r$ a resistor??? IDK, it's strange to see two batteries in series if they aren't modeling something about two different batteries. It's even stranger to use that schematic symbol for a resistor. I'd guess it's a battery (ideal voltage source, really).

 

[Steve4Physics]

Now , by removing the battery $\epsilon$ , i got $R_{\text{th}}=\frac{5R}{3}+r$

You don't remove the battery, you short-circuit it, i.e. replace it with a wire. Then you find the resistance between the load terminals. Check you have done this correctly.

Edit. To be more precise, I should have said short-circuit the emf: the battery's internal resisstance should still be part of the circuit - it doesn't get shorted-out.

 

[Bling Fizikst]

You don't remove the battery, you short-circuit it, i.e. replace it with a wire. Then you find the resistance between the load terminals. Check you have done this correctly.

Yeah!!! i made that silly error of replacing the battery with open circuit . I have arrived at the correct answer now . Thanks!