Question

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 $E_0$? Explain.

Short answer

a. Resonance frequency in rad/s $=3125rad/s$

Resonance frequency in $Hz=500Hz$

b. $V_R=10V$; $V_L=31.25V$

c.$V_c$and$V_L$have opposite phases

Step-by-step solution

Part a) step 1 : Given information 

We have to find the resonance frequency

Detailed solution 

In a series LCR Circuit, the resonance frequency $f_R$(in Hz) is given by

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

It is given that $L=10mH,C=10\mu F$

Thus,

$$\begin{gathered} f_R=\frac{1}{2\pi \sqrt{(10\times {10}^{-3}H)(10\times {10}^{-6}}F)}=\frac{1}{2\pi \sqrt{{10}^{-7}}}=\frac{{10}^3}{2\pi \times 0.32} \\ \Rightarrow f_R=500Hz \end{gathered}$$

Now,

$\Rightarrow$$$\begin{gathered} {\omega }_r(rad/s)=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10\times {10}^{-3}\times 10\times {10}^{-6}}}=\frac{1}{\sqrt{{10}^{-7}}} \\ {\omega }_r=3125rad/s \end{gathered}$$

part b)  step 1 : Given Information 

We have to find $V_R$and $V_L$at resonance.

part b) Step 2 : Detailed solution 

As we know , Current = Voltage / Impedence

$$\begin{gathered} I=\frac{V}{\sqrt{R^2+(\omega L-{\displaystyle \frac{1}{\omega C}}}{)}^2}=\frac{10}{\sqrt{{10}^2+{(3125\times 10\times {10}^{-3}-{\displaystyle \frac{1}{3125\times 10\times {10}^{-6}}})}^2}} \\ \cong 0.998A=1A \end{gathered}$$

Now on putting values in respective formulae;

$$\begin{gathered} V_L=I{X}_L=IL\omega \\ =1\times {10}^{-3}\times 3125 \\ =3125\times {10}^{-2}=31.25V \\ {V}_R=IR=1\times 10=10V \end{gathered}$$

Part c) 

The voltage drop across the inductor can be greater than the applied voltage of the a.c because $V_C$and $V_L$have opposite Phases . Thus,$V_C$ or $V_L$may be greater than V or $E_0$.