91Ó°ÊÓ

We have been given that we need to show the average power loss in a series RLC circuit is

$P_{avg}=\frac{{\mathrm{Ó\neg }}^2{\left({\mathrm{Î\mu }}_{rms}\right)}^{2}R}{{\mathrm{Ó\neg }}^2{R}^2+L^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2}$

We know that the power dissipated ($P_{avg}$) is given by:

$P_{avg}=I_{rms}{\mathrm{Î\mu }}_{rms}\cos \left(\mathrm{Ï•}\right)$ (Let this equation be$\left(1\right)$)

where, $I_{rms}$is root mean square current, ${\mathrm{Î\mu }}_{rms}$is root mean square voltage and $\cos \left(\mathrm{Ï•}\right)$is the power factor.

We know the formulas,

$I_{rms}=\frac{{\mathrm{Î\mu }}_{rms}}{Z}$and $\cos \left(\mathrm{Ï•}\right)=\frac{R}{Z}$ ,

where $R$is the resistance in Ohms ($\mathrm{Î\copyright }$) and $Z$is impedance.

Substituting the values of $I_{rms}$and $\cos \left(\mathrm{Ï•}\right)$in equation ($1$), we get:

$P_{avg}=\frac{{\mathrm{Î\mu }}_{rms}^{2}R}{Z^2}$ (Let this equation be$\left(2\right)$)

We know the formula to calculate $Z^2$:

$Z^2=R^2+{(X_L-X_C)}^2$,

where $X_L$is the inductive reactance and $X_C$is the capacitive reactance.

Substituting$X_L=\mathrm{Ó\neg }L$andlocalid="1650341668142" $X_C=\frac{1}{\mathrm{Ó\neg }C}$, we get:

$Z^2=R^2+{\left(\mathrm{Ó\neg }L-\frac{1}{\mathrm{Ó\neg }C}\right)}^2$

$Z^2=R^2+\frac{L^2}{{\mathrm{Ó\neg }}^2}{\left({\mathrm{Ó\neg }}^2-\frac{1}{LC}\right)}^2$

Substituting $\frac{1}{LC}={\mathrm{Ó\neg }}_o^2$, we get:

$Z^2=R^2+\frac{L^2}{{\mathrm{Ó\neg }}^2}{\left({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2\right)}^2$

Now, substitute the value of $Z^2$in equation $\left(2\right)$to get the expression of $P_{avg}$.

$P_{avg}=\frac{{\mathrm{Î\mu }}_{rms}^{2}R}{R^2+{\displaystyle \frac{L^2}{{\mathrm{Ó\neg }}^2}}{\left({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2\right)}^2}$

$P_{avg}=\frac{{\mathrm{Ó\neg }}^2{\left({\mathrm{Î\mu }}_{rms}\right)}^{2}R}{{\mathrm{Ó\neg }}^2{R}^2+L^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2}$

Hence Proved.

We have been given that we need to prove the energy dissipation is a maximum at $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_o$.

Energy dissipation is maximum when $\frac{d{P}_{avg}}{d\mathrm{Ó\neg }}=0$.

Now simplify the derivative of $P_{avg}$obtained in Part(a) with respect to $\mathrm{Ó\neg }$:

$\frac{d{P}_{avg}}{d\mathrm{Ó\neg }}=0$

role="math" localid="1650343482848" $\frac{d\left({\displaystyle \frac{{\mathrm{Ó\neg }}^2{\left({\mathrm{Î\mu }}_{rms}\right)}^{2}R}{{\mathrm{Ó\neg }}^2{R}^2+L^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2}}\right)}{d\mathrm{Ó\neg }}=0$

role="math" localid="1650343502884" $\frac{2{\mathrm{Î\mu }}_{rms}^{2}R\mathrm{Ó\neg }\left(R^2{\mathrm{Ó\neg }}^2+L^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2\right)-{\mathrm{Î\mu }}_{rms}^{2}R{\mathrm{Ó\neg }}^2\left(2\mathrm{Ó\neg }{R}^2+2L^2\left({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2\right)2\mathrm{Ó\neg }\right)}{{\left(R^2{\mathrm{Ó\neg }}^2+L^2{\left({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2\right)}^2\right)}^2}=0$

On simplifying, we get:

$2{\mathrm{Î\mu }}_{rms}^2{R}^3{\mathrm{Ó\neg }}^3+2{\mathrm{Î\mu }}_{rms}^{2}R\mathrm{Ó\neg }{L}^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2-2{\mathrm{Î\mu }}_{rms}^2{R}^3{\mathrm{Ó\neg }}^3-4{\mathrm{Î\mu }}_{rms}^{2}R{\mathrm{Ó\neg }}^3{L}^2({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)=0$

$2{\mathrm{Î\mu }}_{rms}^{2}R\mathrm{Ó\neg }{L}^2{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2-4{\mathrm{Î\mu }}_{rms}^{2}R{\mathrm{Ó\neg }}^3{L}^2({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)=0$

$2{\mathrm{Î\mu }}_{rms}^{2}R\mathrm{Ó\neg }{L}^2\left[{({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2-2{\mathrm{Ó\neg }}^2({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)\right]=0$

${({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)}^2-2{\mathrm{Ó\neg }}^2({\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_o^2)=0$

From the previous expression, we can observe a relation which is solved by $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_o$.

Hence, we have proved that the power is maximized when $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_o$.