Question

You have a 200Ω resistor, a 0.400-H inductor, a 5.00mF capacitor, and a variable frequency ac source with an amplitude of 3.00 V. You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of 400 rad/s? At this frequency, will the source voltage lead or lag the current?

Short answer

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

Step-by-step solution

Step-1: Formulas used  

Resonance frequency is given by${\mathbf{f}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\tau \tau LC}}$.The reactance are given by${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$ for capacitor.

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

Step-2: Calculations for resonance frequency and corresponding current amplitude.

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

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

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

Step-3: Calculations for current amplitude at an angular frequency of 400 rad/s.

First calculate the inductive reactance;

$$\begin{gathered} {\mathrm{X}}_{\mathrm{L}}=2\mathrm{\tau \tau }\left(113\mathrm{Hz}\right)\left(0.4\mathrm{H}\right) \\ =160\mathrm{\Omega } \end{gathered}$$

And then the capacitive reactance;

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{\tau \tau }\left(113\mathrm{Hz}\right)\left(5\times {10}^{-6}\mathrm{F}\right)} \\ =500\mathrm{\Omega } \end{gathered}$$

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

Then calculate the impedance;

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

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

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

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.