Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω.

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Here, we are given a LCR circuit. 

Inductance, L = 5.0 H
Resistance, R = 40 $\mathrm{\Omega }$
Capacitance, $\mathrm{C} = 80 \mathrm{\mu F} = 80 \times {10}^{-6}\mathrm{F}$
Effective voltage, ${\mathrm{E}}_{\mathrm{v}} = 230 \mathrm{volt}$ 
$\Rightarrow$Peak voltage,  ${\mathrm{E}}_0 = \sqrt{2} {\mathrm{E}}_{\mathrm{v}} = \sqrt{2} \times 230 \mathrm{V}$ 

(a) Resonance angular frequency is given by,
                $$\begin{gathered} {\mathrm{\omega }}_{\mathrm{r}} = \frac{1}{\sqrt{\mathrm{LC}}} \\ = \frac{1}{\sqrt{5\times 80\times {10}^{-6}}} \\ = \frac{1}{2 \times {10}^{-2}} \\ = 50 \mathrm{rad}/\sec . \end{gathered}$$ 

(b) Impedance of the circuit,

              $\mathrm{Z} = \sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L} -\frac{1}{\mathrm{\omega C}}\right)}^2}$

At resonance,   $\mathrm{\omega L} = \frac{1}{\mathrm{\omega C}}$
Therefore, 

$\mathrm{Z} = \sqrt{{\mathrm{R}}^2} = \mathrm{R} = 40 \mathrm{\Omega }$ 

Amplitude of current at resonating frequency

$$\begin{gathered} \mathrm{Peak} \mathrm{value} \mathrm{of} \mathrm{current}, {\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\mathrm{z}} = \frac{\sqrt{2}\times 230}{40} = 8.13 \mathrm{A} \\ \mathrm{Rms} \mathrm{value} \mathrm{of} \mathrm{current}, {\mathrm{I}}_{\mathrm{v}} = \frac{{\mathrm{I}}_0}{\sqrt{2}} = \frac{8.13}{\sqrt{2}} = 5.75 \mathrm{A} \end{gathered}$$

(c) Potential drop across L

$$\begin{gathered} {\mathrm{V}}_{\mathrm{L} \mathrm{rms}} = {\mathrm{I}}_{\mathrm{v}} {\mathrm{\omega }}_{\mathrm{r}} \mathrm{L} \\ = 5.75 \times 50 \times 5.0 \\ = 1437.5 \mathrm{V} \end{gathered}$$ 

Potential drop across R

 $$\begin{gathered} {\mathrm{V}}_{\mathrm{R} \mathrm{rms}} = {\mathrm{I}}_{\mathrm{v}} \times \mathrm{R} \\ = 5.75 \times 40 \\ = 230 \mathrm{volt} \end{gathered}$$ 

Potential drop across C

 ${\mathrm{V}}_{\mathrm{C} \mathrm{rms}} = {\mathrm{I}}_{\mathrm{v}} \left(\frac{1}{{\mathrm{\omega }}_{\mathrm{r}}\mathrm{C}}\right)$
          $$\begin{gathered} = 5.75 \times \frac{1}{50 \times 80 \times {10}^{-6}} \\ = \frac{5.75}{4}\times {10}^3 \\ = 1437.5 \mathrm{V} \end{gathered}$$ 

Therefore, 

Potential drop across LC circuit 

${\mathrm{V}}_{\mathrm{LC} \mathrm{rms}} = {\mathrm{V}}_{\mathrm{L} \mathrm{rms}} - {\mathrm{V}}_{\mathrm{C} \mathrm{rms}} = 0$ 

Thus, the potential drop across the LC combination is zero at the resonating frequency.