For the circuit of FIGURE EX32.32,

a. What is the resonance frequency, in both rad/s and Hz?

b .Find $V_R$and $V_L$at resonance.

c. How can $V_L$be larger than ${\epsilon }_o$? Explain.

We have been given values of

Inductance ($L$)= $10mH$

Capacitance($C$) =$10\mu F$

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

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

$f_r=\frac{1}{2\pi \sqrt{LC}}$

Given ; $L=10mH,C=10\mu F$

$$\begin{gathered} f_r=\frac{1}{2\pi \sqrt{10\times {10}^{-3}\times 10\times {10}^{-6}}} \\ =\frac{1}{2\pi \sqrt{{10}^{-7}}} \\ =\frac{{10}^3}{2\pi \times 0.32}=\frac{1000}{2} \\ \Rightarrow f_r=500Hz \end{gathered}$$

Now, $$\begin{gathered} {\omega }_r=2{\mathrm{\pi f}}_{\mathrm{r}} \\ \Rightarrow {\omega }_r=3140rad/s \end{gathered}$$

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

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

${\omega }_r=3140rad/s$

Voltage = $10V$

$R=10\Omega ;L=10mH;C=10\mu F$

$$\begin{gathered} I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+(\omega L-{\displaystyle \frac{1}{\omega C}}}{)}^2} \\ =\frac{10}{\sqrt{{10}^2+(3140\times 10\times {10}^{-3}-{\displaystyle \frac{1}{3140\times 10\times {10}^{-6}}}}{)}^2}\cong 0.998=1A \\ \Rightarrow V_L=X_{L}I=L\omega I \\ \Rightarrow V_L=10\times {10}^{-3}\times 3140\times 1 \\ \Rightarrow V_L=31.40V \\ {V}_R=IR=1\times 10 \\ \Rightarrow V_R=10V \end{gathered}$$

We need to explain how$V_L$can be larger than${\epsilon }_o$

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${\epsilon }_o$