91Ó°ÊÓ

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

We Know that,

Denominator equal to zero

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

$-\frac{2}{{\mathrm{Ó\neg }}^3}\frac{R^2}{h^2}+2\left(1-\frac{1}{{\mathrm{Ó\neg }}^{2}hc}\right)×\left(0+\left(\frac{2}{{\mathrm{Ó\neg }}^3}\right)\left(\frac{1}{LC}\right)\right)=0$

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

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

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

The voltage across the inductor is

$V_L=I{X}_L=\frac{{\mathrm{Î\mu }}_0}{Z}{X}_L=\frac{{\mathrm{Î\mu }}_0}{Z}\mathrm{Ó\neg }L$

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

$V_L=\frac{{\mathrm{Î\mu }}_0}{Z}{\mathrm{Ó\neg }}_{0}L=\frac{{\mathrm{Î\mu }}_0}{Z}\sqrt{\frac{L}{C}}=\frac{10V}{1.0ohm}\sqrt{\frac{1.0\mathrm{μ}H}{1.0\mathrm{μ}F}}=10V$

The maximizing frequency ${\mathrm{Ó\neg }}_L$is

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

The impedance at this frequency is

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

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

The Voltage is

$V_L=\frac{{\mathrm{Î\mu }}_0}{Z}{\mathrm{Ó\neg }}_{L}L=\frac{10.0V}{1.225ohm}(1.414×{10}^{6}rad/s)(1.0\mathrm{μ}H)=11.55V≈12V$