Question

An L-R-C series circuit is constructed using a 175Ωresistor, a 12.5mF capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

Short answer

a)The angular frequency at which the impedance will be the smallest is $\omega =3162\frac{rad}{s}$.

b)The maximum current through the inductor is 143 mA.c.

c)The potential difference across the ac source is ++, the resistor is $V_R=12.5V$, the capacitor isrole="math" localid="1668252349319" $V_C=+12.5V$and the inductor is $V_C=+3.13V$.

d)The summation of potential differences equals the voltage amplitude.

Step-by-step solution

Step-1: Formulas used  

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

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

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IZ

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

V=IX

Where X is the reactance which is equal to${\mathit{X}}_{\mathbf{L}}\mathbf{=}\mathit{\omega }\mathit{L}$ for inductor and${\mathit{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega }\mathbf{C}}$ for capacitor and Rfor resistor.

The process of peaking the current at a particular frequency represents the resonance and that frequency is known as resonant frequency.

$\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{L}\mathbf{C}}}$,whereis the resonant angular frequency.

At resonance,$X_L=X_C$.

At resonance, the impedance is minimum and this minimum value is same as that of resistor.

The current of the AC Source is

$\mathbf{i}\mathbf{=}\mathbf{lcos}\left(\mathrm{\omega t}\right)$where t is time and is the angular frequency.

The Voltage of the AC Source is

$\mathbf{v}\mathbf{=}\mathbf{Vcos}\left(\mathrm{\omega t}\right)$

The voltages across R, C and L is

$\begin{array}{r}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{I}\mathbf{R}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\left(}\mathbf{\omega }\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ {\mathbf{V}}_{\mathbf{C}}\mathbf{=}{\mathbf{I}}_{\mathbf{C}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\left(\omega t-{90}^{\circ }\right)\mathbf{}\mathbf{}\\ {\mathbf{V}}_{\mathbf{L}}\mathbf{=}\mathbf{I}{\mathbf{X}}_{\mathbf{L}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\left(\omega t+{90}^{\circ }\right)\end{array}\mathbf{}\mathbf{}\mathbf{}$

Step-2: Calculations for resonant angular frequency and impedance.

$\begin{array}{r}\omega =\frac{1}{\sqrt{LC}}\\ \begin{array}{rl}L& =8\mathrm{mH}\\ C& =12.5\mu \mathrm{F}\\ \omega & =\frac{1}{\sqrt{\left(8\times {10}^{-3}\mathrm{H}\right)\left(12.5\times {10}^{-8}\mathrm{C}\right)}}\\ & =3162\frac{\mathrm{rad}}{\mathrm{s}}\end{array}\end{array}$

At resonance,

$$\begin{gathered} Z=R \\ =175\Omega \end{gathered}$$

Step-3: Calculations for maximum current through the circuit

$\begin{array}{r}\mathrm{I}=\frac{\mathrm{V}}{Z}\\ \mathrm{V}=25\mathrm{V}\\ \mathrm{Z}=175\mathrm{\Omega }\\ \mathrm{I}=\frac{25\mathrm{V}}{175\mathrm{\Omega }}\\ =143\mathrm{mA}\end{array}$

Step-4: Calculations for voltage amplitude across resistor, capacitor,inductor and source.

At time t, the current I is half the amplitude current I.

$\begin{array}{r}\mathrm{i}=\frac{1}{2}\\ \mid \cos \omega t=\frac{1}{2}\\ \cos \omega t=\frac{1}{2}\\ \omega t={60}^{\circ }\end{array}$

Using,

$$\begin{gathered} \begin{array}{r}{V}_R=IR\cos \left(\omega t\right)\\ V_C=I{X}_c\cos \left(\omega t-{90}^{\circ }\right)\\ V_L=I{X}_L\cos \left(\omega t+{90}^{\circ }\right)\\ V_R=\left(0.143\mathrm{A}\right)\left(175\mathrm{\Omega }\right)\left(0.5\right)\\ =12.5\mathrm{V}\\ V_C=\left(0.143\mathrm{A}\right)\frac{1}{\left(3162\frac{\mathrm{rad}}{\mathrm{s}}\right)\left(12.5\times {10}^{-6}\mathrm{F}\right)}\cos \left({60}^{\circ }-{90}^{\circ }\right)\\ =+3.13\mathrm{V}\\ V_L=\left(0.143\mathrm{A}\right)\left(3162\frac{\mathrm{rad}}{\mathrm{s}}\right)\left(8\times {10}^{-3}\right)\left(\cos \left({60}^{\circ }+{90}^{\circ }\right)\right.\\ =-3.13\mathrm{V}\\ \mathrm{V}=\mathrm{Vcos}\left(\omega \mathrm{t}\right)\end{array} \\ v=\left(25V\right)(0.5) \\ =12.5V \end{gathered}$$

Step-5: Relation voltage amplitude across resistor, capacitor, inductor and source.

The summation of the voltages across resistor, capacitor and inductors equals the source voltage.

Therefore, a)The angular frequency at which the impedance will be the smallest is $\omega =3162\frac{rad}{s}$.b)The maximum current through the inductor is 143mA .c)The potential difference across the ac source is 12.5V, the resistor is $V_R=12.5V$, the capacitor is $V_C=+3.13V$and the inductor is $V_L=-3.13V$.d)The summation of potential differences equals the voltage amplitude.