91Ó°ÊÓ

Resonance frequency is given by${\mathbf{f}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï„Ï„³¢°ä}}$.The reactance are given by${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{Ó\neg ³¢}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{Ó\neg °ä}}$ for capacitor.

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_{C)}\right)}^{\mathbf{2}}}$,where Z is the impedance

$$\begin{gathered} {\mathrm{f}}_0=\frac{1}{2\mathrm{ττ}\left(0.4\mathrm{H}\right)\left(5×{10}^{-6}\mathrm{F}\right)} \\ =113\mathrm{Hz} \end{gathered}$$

For the current, use Ohm’s law to get the current.

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ =\frac{3\mathrm{V}}{200\mathrm{Î\copyright }} \\ =15\mathrm{mA} \end{gathered}$$

First calculate the inductive reactance;

$$\begin{gathered} {\mathrm{X}}_{\mathrm{L}}=2\mathrm{Ï„Ï„}\left(113\mathrm{Hz}\right)\left(0.4\mathrm{H}\right) \\ =160\mathrm{Î\copyright } \end{gathered}$$

And then the capacitive reactance;

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{Ï„Ï„}\left(113\mathrm{Hz}\right)\left(5×{10}^{-6}\mathrm{F}\right)} \\ =500\mathrm{Î\copyright } \end{gathered}$$

Since the capacitive reactance is greater than the inductive reactance, therefore the voltage will lag behind the current.

Then calculate the impedance;

$$\begin{gathered} \mathrm{Z}=\sqrt{{200}^2+{\left(160-500\right)}^2} \\ =394.5\mathrm{Î\copyright } \end{gathered}$$

The impedance represents the total resistance in the circuit, so we can use its value in Ohm’s law to get the current amplitude.

$$\begin{gathered} \mathrm{I}=\frac{3\mathrm{V}}{394.5\mathrm{Î\copyright }} \\ =7.6\mathrm{mA} \end{gathered}$$

Therefore (a)The current will be greatest at a frequency of 113Hz, current amplitude at this frequency is 15mA and (b).At 400rad/s frequency, current amplitude will be 7.6mA. The source voltage will lag the current.