Question

a. Show that the peak inductor voltage in a series RLC circuit is maximum at frequency

${\omega }_L={\left(\frac{1}{{\omega }_o^2}-\frac{1}{2}{R}^2{C}^2\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

b. A series RLC circuit with ${\epsilon }_o=10.0V$consists of a $1.0\Omega$resistor, a $1.0\mu H$inductor, and a $1.0\mu F$capacitor. What is $V_L$at $\omega ={\omega }_o$and $\omega ={\omega }_L$?

Short answer

a. The peak inductor voltage in a series RLC circuit is maximum at frequency is$\omega =\frac{w_o}{\sqrt{1-\frac{R^{2}C}{2T}}}$

b.$V_L$at$\omega ={\omega }_0$is$10V$and at$\omega ={\omega }_L$is$12V$

Step-by-step solution

Part (a) Step 1: Given information

We have to prove that the peak inductor voltage in a series RLC circuit is maximum at frequency $\omega =\frac{w_o}{\sqrt{1-\frac{R^{2}C}{2T}}}$

Part (a) Step 2: Simplification

We Know that,

$$\begin{gathered} V_L=\frac{V_s\omega L}{\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}} \\ {V}_L=\frac{1}{\sqrt{\frac{R^2}{{\omega }^2{L}^2}+{\left(1-\frac{1}{{\omega }^{2}LC}\right)}^2}} \end{gathered}$$

Denominator equal to zero

$\frac{d}{d\omega }\left\{\frac{R^2}{w^2{h}^2}+{\left(1-\frac{1}{w^{2}kC}\right)}^2\right\}=0$

$-\frac{2}{{\omega }^3}\frac{R^2}{h^2}+2\left(1-\frac{1}{{\omega }^{2}hc}\right)\times \left(0+\left(\frac{2}{{\omega }^3}\right)\left(\frac{1}{LC}\right)\right)=0$

We know that $w_o^2=\frac{1}{LC}$

$$\begin{gathered} -\frac{2}{w^3}\frac{R^2}{h^2}+4{\left(1-\frac{{\omega }_o^2}{{\omega }^2}\right)}^2\frac{{\omega }_o^2}{{\omega }^3}=0 \\ 1-\frac{{\omega }_o^2}{w^2}=\frac{R^{2}C}{2L} \\ \omega =\frac{w_o}{\sqrt{1-\frac{R^{2}C}{2L}}} \end{gathered}$$

Part (b) Step 1 : Given Information

We have given that a series RLC circuit with ${\epsilon }_0=10.0V$consists of a $1.0ohm$resistor, a $1.0\mu H$inductor, and a $1.0\mu F$capacitor we have to find$V_L$at$\omega ={\omega }_0$and$\omega ={\omega }_L$.

Part (b) Step 2 : Simplification

The voltage across the inductor is

$V_L=I{X}_L=\frac{{\epsilon }_0}{Z}{X}_L=\frac{{\epsilon }_0}{Z}\omega L$

At $\omega ={\omega }_0=\frac{1}{\sqrt{LC}},Z=R=1.0ohm.$We get,

$V_L=\frac{{\epsilon }_0}{Z}{\omega }_{0}L=\frac{{\epsilon }_0}{Z}\sqrt{\frac{L}{C}}=\frac{10V}{1.0ohm}\sqrt{\frac{1.0\mu H}{1.0\mu F}}=10V$

The maximizing frequency ${\omega }_L$is

$$\begin{gathered} {\omega }_L={\left(\frac{1}{{\omega }_o^2}-\frac{1}{2}{R}^2{C}^2\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(LC-\frac{1}{2}{R}^2{C}^2\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\omega }_L={\left((1.0\mu H)(1.0\mu F)-\frac{1}{2}{(1.0ohm)}^2{(1.0\mu F)}^2\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.414\times {10}^{6}rad/s \end{gathered}$$

The impedance at this frequency is

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

$Z=\sqrt{{(1.0)}^2+{\left((1.414\times 106)(1.0\mu H)-\frac{1}{(1.414\times 106)(1.0\mu F)}\right)}^2}=1.225ohm$

The Voltage is

$V_L=\frac{{\epsilon }_0}{Z}{\omega }_{L}L=\frac{10.0V}{1.225ohm}(1.414\times {10}^{6}rad/s)(1.0\mu H)=11.55V\approx 12V$