IIT JEE , AIEEE Forum - Physics , Mechanics

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Proof for purely resistive circuits

In the diagram opposite, power is being transferred from the source, with voltage

V

and fixed source resistance

R S

, to a load with resistance

R L

, resulting in a current

I

. By Ohm's law,

I

is simply the source voltage divided by the total circuit resistance:

$I = {V over R_mathrm{S} + R_mathrm{L}}.$

The power

P L

dissipated in the load is the square of the current multiplied by the resistance:

$P_mathrm{L} = I^2 R_mathrm{L} = {{ left( {V over {R_mathrm{S} + R_mathrm{L}}} ight) }^2} R_mathrm{L} = {{V^2} over {R_mathrm{S}^2 / R_mathrm{L} + 2R_mathrm{S} + R_mathrm{L}}}.$

We could calculate the value of

R L

for which this expression is a maximum, but it is easier to calculate the value of

R L

for which the denominator

$R_mathrm{S}^2 / R_mathrm{L} + 2R_mathrm{S} + R_mathrm{L}$

is a minimum. The result will be the same in either case. Differentiating with respect to

R L

:

${dover{dR_mathrm{L}}} left( {R_mathrm{S}^2 / R_mathrm{L} + 2R_mathrm{S} + R_mathrm{L}} ight) = -R_mathrm{S}^2 / R_mathrm{L}^2+1.$

For a maximum or minimum, the first derivative is zero, so

${R_mathrm{S}^2 / R_mathrm{L}^2} = 1$

or

$R_mathrm{L} = pm R_mathrm{S}.$

In practical resistive circuits,

R S

and

R L

are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:

${{d^2} over {dR_mathrm{L}^2}} left( {R_mathrm{S}^2 / R_mathrm{L} + 2 R_mathrm{S} + R_mathrm{L}} ight) = {2 R_mathrm{S}^2} / {R_mathrm{L}^3}$

This is positive for positive values of

R S

and

R L

, showing that the denominator is a minimum, and the power is therefore a maximum, when

R S = R L .

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