Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.

Given, an LCR circuit. L, C and R are arranged in parallel and the source frequency is kept equal to the resonating frequency.

Then,

Using the formula for resonant frequency,
$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{r}} = \frac{1}{\sqrt{\mathrm{LC}}} \\ = \frac{1}{\sqrt{5 \times 80\times {10}^{-6}}} \\ = \frac{1}{\sqrt{400 \times {10}^{-6}}} \\ = 50 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$ 

Since elements are in parallel, reactance X of L and C in parallel is given by 

$\frac{1}{\mathrm{X}} = \frac{1}{\mathrm{\omega L}} = \frac{1}{1/\mathrm{\omega C}}= \frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}$ 

Impedance of R and X in parallel is given by
                                 $\frac{1}{\mathrm{Z}} = \sqrt{\frac{1}{{\mathrm{R}}^2}+\frac{1}{{\mathrm{X}}^2}} = \sqrt{\frac{1}{{\mathrm{R}}^2}+{\left(\frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}\right)}^2}$
                    $\frac{1}{\mathrm{Z}} = \sqrt{\frac{1+{\mathrm{R}}^2{\left({\displaystyle \frac{1}{\mathrm{\omega L}}}-\mathrm{\omega C}\right)}^2}{\mathrm{R}}}$  

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

which is less than resistance R.  

At resonant frequency,
                                      $\mathrm{\omega L} = \frac{1}{\mathrm{\omega C}} \mathrm{or} \mathrm{\omega C} = \frac{1}{\mathrm{\omega L}}$ 

and       $\left(\frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}\right) = 0$ 

Then, impedance Z = R and will be maximum.

Hence, current will be minimum at resonant frequency in the parallel LCR circuit.

From Ex. 11: 

Inductance, L = 5H
Capacitance, C = 80 × 10^–6 F
Resistnace, R = 40 Ω.
 ${\mathrm{E}}_{\mathrm{rms}} = 230 \mathrm{V}.$ 

Therfore,

$$\begin{gathered} {\left({\mathrm{I}}_{\mathrm{R}}\right)}_{\mathrm{rms}} = \frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{R}} = \frac{230}{40} = 5.75 \mathrm{A}. \\ ({\mathrm{I}}_{\mathrm{L}}{)}_{\mathrm{rms}} = \frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{\omega L}} = \frac{230}{50 \times 5} = 0.92 \mathrm{A}. \\ ({\mathrm{I}}_{\mathrm{C}}{)}_{\mathrm{rms}} = \frac{{\mathrm{V}}_{\mathrm{rms}}}{1/\mathrm{\omega C}} = 230 \times 50 \times 80 \times {10}^{-6} = 0.92 \mathrm{A}. \end{gathered}$$

Current through L and C will be in opposite phase. Hence, current in circuit will be only 5.75 A$\left(= \frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{R}}\right)$ as, circuit impedance will be equal to R only.