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Maximum Power is transferred to a load when load resistance is equal to internal resistance of source. I want to know How ?

If I consider a source with internal resistance 'R' ,connected with resistance 'R' and current through circuit is 'i'. Then power is P1=$ (i^2)$2R. But if this source is connected to resistance 5R i.e internal resistance not equal to external resistance then Power P2= $(i^2)$6R, which is greater than P1. Then how the given statement above is true.

Avi
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  • Bit of advice: Don't use the ? punctuation unnecessarily like ??? ; also, this is a MathJax-enabled site; you can use the facility to format your equations; check this Meta Math.SE post for a quick review of MathJax. –  Dec 25 '16 at 07:39

1 Answers1

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If the voltage of the source is $V$ then the current $i= \dfrac{V}{R+r}$ where $r$ is the source resistance and $R$ is the load resistance.

So the power dissipated in the load resistance $R$ is $\dfrac{V^2}{(R+r)^2}R$.

Your error is to assume that the current $i$ does not change and try to evaluate the power dissipated in both resistors.

Substituting $r=R$ and $r=\dfrac R5$ will show you that the $r=R$ situation dissipates more power in the load resistor $R$ than when $r=\dfrac R5$.

Farcher
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