A series RLC circuit consists of a $50\Omega$ resistor, a localid="1650177582139" $3.3mH$ inductor, and a localid="1650177586647" $480nF$ capacitor. It is connected to an oscillator with a peak voltage of localid="1650177591401" $5.0V$. Determine the impedance, the peak current, and the phase angle at frequencies :

(a) localid="1650177630395" $3000Hz$(b) localid="1650177635339" $4000Hz$and (c) localid="1650177638990" $5000Hz.$

We have been given that

Resistance $\left(R\right)=50\Omega$

Inductance $\left(\mathrm{L}\right)=3.3mH=3.3\times {10}^{-3}H$

Capacitance $\left(C\right)=480\eta F=480\times {10}^{-9}F$

Peak voltage $\left(V_o\right)=5V$

We have to find impedence , peak current and phase angle at given frequency

We know that

$Z=\sqrt{{\left(50\right)}^2+{(X_L-X_C)}^2}$ where $f=3000Hz$

$$\begin{gathered} X_L=\omega L=2\times \mathrm{\pi }\times 3000\times 3.3\times {10}^{-3} \\ {\mathrm{X}}_{\mathrm{L}}\simeq 62\mathrm{\Omega } \end{gathered}$$

localid="1650178194248" $X_C=\frac{1}{2\pi fC}=111\Omega$

$$\begin{gathered} \Rightarrow Z=\sqrt{2500+{(111-62)}^2} \\ \Rightarrow Z=70\Omega \end{gathered}$$

Also,

$I=\frac{V_{max}}{Z}=\frac{5}{70}=0.071A$

Finding phase angle,

$$\begin{gathered} \tan \varphi =\frac{X_L-X_C}{R}=\frac{111-62}{50}=0.98\approx 1 \\ \Rightarrow \varphi =\frac{\pi }{4} \end{gathered}$$

We are given: Frequency$\left(f\right)=4000Hz$

We have to find impedance , peak current and phase angle at given frequency.

$$\begin{gathered} Z=\sqrt{{\left(50\right)}^2+{(X_L-X_C)}^2} \\ {X}_L=2\pi fL=83\Omega ;X_c=\frac{1}{2\pi fC}=83\Omega \end{gathered}$$

Where $f=4000Hz$

$Z=\sqrt{2500}=50\Omega$

Now, Phase angle ;localid="1650177774479" $$\begin{gathered} I=\frac{V_{max}}{Z}=\frac{5}{50}=0.1A \\ \tan \varphi =\frac{X_L-X_C}{R}=0 \\ \Rightarrow \varphi =0,\pi \end{gathered}$$

We are given: Frequency $\left(f\right)=5000Hz$

We have to find impedance , peak current and phase angle at given frequency.

$$\begin{gathered} f=5000Hz \\ {X}_L=2\times 3.14\times 5000\times 3.3\times {10}^{-3}=104\Omega \\ {X}_C=\frac{1}{2\pi fC}=67\Omega \\ \Rightarrow Z=\sqrt{2500+{(104-67)}^2}=62\Omega \\ {I}_m=\frac{V_{max}}{Z}=\frac{5}{62}=0.08A \\ \tan \varphi =\frac{(104-67)}{50}=0.74 \\ \varphi =36.5° \end{gathered}$$