91Ó°ÊÓ

We have been given values of

Inductance ($L$)= $10mH$

Capacitance($C$) =$10\mathrm{μ}F$

We need to find out the resonant frequency , $f_r$(in Hz) and ${\mathrm{Ó\neg }}_r$(in rad/s)

In a series LCR Circuit, the resonant frequency ,$f_r$is given by

$f_r=\frac{1}{2\mathrm{Ï€}\sqrt{LC}}$

Given ; $L=10mH,C=10\mathrm{μ}F$

$$\begin{gathered} f_r=\frac{1}{2\mathrm{π}\sqrt{10×{10}^{-3}×10×{10}^{-6}}} \\ =\frac{1}{2\mathrm{π}\sqrt{{10}^{-7}}} \\ =\frac{{10}^3}{2\mathrm{π}×0.32}=\frac{1000}{2} \\ ⇒f_r=500Hz \end{gathered}$$

Now, $$\begin{gathered} {\mathrm{Ó\neg }}_r=2{\mathrm{Ï€´Ú}}_{\mathrm{r}} \\ ⇒{\mathrm{Ó\neg }}_r=3140rad/s \end{gathered}$$

We need to find out $V_R$and $V_L$at resonance

Resonance frequency ($f_r$) = $500Hz$

${\mathrm{Ó\neg }}_r=3140rad/s$

Voltage = $10V$

$R=10\mathrm{Î\copyright };L=10mH;C=10\mathrm{μ}F$

$$\begin{gathered} I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+(\mathrm{Ó\neg }L-{\displaystyle \frac{1}{\mathrm{Ó\neg }C}}}{)}^2} \\ =\frac{10}{\sqrt{{10}^2+(3140×10×{10}^{-3}-{\displaystyle \frac{1}{3140×10×{10}^{-6}}}}{)}^2}â‰\dots 0.998=1A \\ ⇒V_L=X_{L}I=L\mathrm{Ó\neg }I \\ ⇒V_L=10×{10}^{-3}×3140×1 \\ ⇒V_L=31.40V \\ {V}_R=IR=1×10 \\ ⇒V_R=10V \end{gathered}$$

We need to explain how$V_L$can be larger than${\mathrm{Î\mu }}_o$

The voltage drop across the inductor can be greater than the applied voltage of ac because as $V_C$and $V_L$have opposite phases. Therefore, $V_C$or$V_L$may be greater than$V$or${\mathrm{Î\mu }}_o$