Question

a. Show that the average power loss in a series RLC circuit is

$P_{avg}=\frac{{\omega }^2{\left({\epsilon }_{rms}\right)}^{2}R}{{\omega }^2{R}^2+L^2{\left({\omega }^2-{\omega }_o^2\right)}^2}$

b. Prove that the energy dissipation is a maximum at$\omega ={\omega }_o$.

Short answer

(a) The proof of the average power loss in a series RLC circuit is given below.

$P_{avg}=\frac{{\omega }^2{\left({\epsilon }_{rms}\right)}^{2}R}{{\omega }^2{R}^2+L^2{({\omega }^2-{\omega }_o^2)}^2}$

(b) The proof of the energy dissipation is a maximum at$\omega ={\omega }_o$is given below.

Step-by-step solution

Part(a) Step 1: Given information

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

$P_{avg}=\frac{{\omega }^2{\left({\epsilon }_{rms}\right)}^{2}R}{{\omega }^2{R}^2+L^2{({\omega }^2-{\omega }_o^2)}^2}$

Part(a) Step 2: Proof

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

$P_{avg}=I_{rms}{\epsilon }_{rms}\cos \left(\varphi \right)$ (Let this equation be$\left(1\right)$)

where, $I_{rms}$is root mean square current, ${\epsilon }_{rms}$is root mean square voltage and $\cos \left(\varphi \right)$is the power factor.

We know the formulas,

$I_{rms}=\frac{{\epsilon }_{rms}}{Z}$and $\cos \left(\varphi \right)=\frac{R}{Z}$ ,

where $R$is the resistance in Ohms ($\Omega$) and $Z$is impedance.

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

$P_{avg}=\frac{{\epsilon }_{rms}^{2}R}{Z^2}$ (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$X_L=\omega L$andlocalid="1650341668142" $X_C=\frac{1}{\omega C}$, we get:

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

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

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

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

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

$P_{avg}=\frac{{\epsilon }_{rms}^{2}R}{R^2+{\displaystyle \frac{L^2}{{\omega }^2}}{\left({\omega }^2-{\omega }_o^2\right)}^2}$

$P_{avg}=\frac{{\omega }^2{\left({\epsilon }_{rms}\right)}^{2}R}{{\omega }^2{R}^2+L^2{({\omega }^2-{\omega }_o^2)}^2}$

Hence Proved.

Part(b) Step 1: Given information

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

Part(b) Step 2: Proof

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

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

$\frac{d{P}_{avg}}{d\omega }=0$

role="math" localid="1650343482848" $\frac{d\left({\displaystyle \frac{{\omega }^2{\left({\epsilon }_{rms}\right)}^{2}R}{{\omega }^2{R}^2+L^2{({\omega }^2-{\omega }_o^2)}^2}}\right)}{d\omega }=0$

role="math" localid="1650343502884" $\frac{2{\epsilon }_{rms}^{2}R\omega \left(R^2{\omega }^2+L^2{({\omega }^2-{\omega }_o^2)}^2\right)-{\epsilon }_{rms}^{2}R{\omega }^2\left(2\omega R^2+2L^2\left({\omega }^2-{\omega }_o^2\right)2\omega \right)}{{\left(R^2{\omega }^2+L^2{\left({\omega }^2-{\omega }_o^2\right)}^2\right)}^2}=0$

On simplifying, we get:

$2{\epsilon }_{rms}^2{R}^3{\omega }^3+2{\epsilon }_{rms}^{2}R\omega L^2{({\omega }^2-{\omega }_o^2)}^2-2{\epsilon }_{rms}^2{R}^3{\omega }^3-4{\epsilon }_{rms}^{2}R{\omega }^3{L}^2({\omega }^2-{\omega }_o^2)=0$

$2{\epsilon }_{rms}^{2}R\omega L^2{({\omega }^2-{\omega }_o^2)}^2-4{\epsilon }_{rms}^{2}R{\omega }^3{L}^2({\omega }^2-{\omega }_o^2)=0$

$2{\epsilon }_{rms}^{2}R\omega L^2\left[{({\omega }^2-{\omega }_o^2)}^2-2{\omega }^2({\omega }^2-{\omega }_o^2)\right]=0$

${({\omega }^2-{\omega }_o^2)}^2-2{\omega }^2({\omega }^2-{\omega }_o^2)=0$

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

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