A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms value.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? ['Average' implies 'averaged over one cycle'.]

Here,
Inductance, L = 80 mH = 80 x 10^-3H 

Capacitance, $\mathrm{C} = 60 \mathrm{\mu F} = 60 \times {10}^{-6}\mathrm{F}$ 
Resistance, R = 0
RMS voltage, E_v = 230 V
$$\begin{gathered} \mathrm{Peak} \mathrm{voltage}, {\mathrm{E}}_0 = \sqrt{2} \times {\mathrm{E}}_{\mathrm{v}} \\ = \sqrt{2} \times 230 \mathrm{V} \end{gathered}$$ 

Frequency of Ac supply,  $\mathrm{f} = 50 \mathrm{Hz}$ 

$$\begin{gathered} \therefore \mathrm{\omega } = 2\mathrm{\pi f} \\ = 100 \mathrm{\pi } \mathrm{rad}/\mathrm{s} \end{gathered}$$ 

(a) We have to find ${\mathrm{I}}_0 = ?, {\mathrm{I}}_{\mathrm{v}} = ?$ 

Therefore,
             $$\begin{gathered} {\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\left(\mathrm{\omega L} -{\displaystyle \frac{1}{\mathrm{\omega L}}}\right)} \\ = \frac{230 \sqrt{2}}{\left(100 \mathrm{\pi } \times 80 \times {10}^{-3} - {\displaystyle \frac{1}{100 \mathrm{\pi } \times 60 \times {10}^{-6}}}\right)} \end{gathered}$$
   $$\begin{gathered} = \frac{230 \sqrt{2}}{\left(8\mathrm{\pi } -{\displaystyle \frac{1000}{6\mathrm{\pi }}}\right)} \\ = \frac{230 \sqrt{2}}{-27.91} \\ = -11.63 \mathrm{amp}. \end{gathered}$$
and,

${\mathrm{I}}_{\mathrm{v}} = \frac{{\mathrm{I}}_0}{\sqrt{2}} =\frac{-11.63}{1.414} = -8.23 \mathrm{amp}.$ 

 Negative sign appears as $\mathrm{\omega L}<\frac{1}{\mathrm{\omega C}}.$ 

$\therefore$ e.m.f lags behind the current by $90°$ 

(b) Rms value of potential drop across L,  

$$\begin{gathered} \mathrm{V} = {\mathrm{I}}_{\mathrm{v}} \times \mathrm{\omega L} \\ = 8.23 \times 100 \mathrm{\pi } \times 80 \times {10}^{-3} \\ = 206.74 \mathrm{volt}. \end{gathered}$$ 

Rms value of potential drop across C,
 $$\begin{gathered} \mathrm{V} = {\mathrm{I}}_{\mathrm{v}} \times \frac{1}{\mathrm{\omega C}} \\ = 8.23 \times \frac{1}{100 \mathrm{\pi } \times 60 \times {10}^{-6}} \\ = 436.84 \mathrm{volt}. \end{gathered}$$ 

As voltages across L and C are 180° out of phase, therefore, they get subtracted.

That is why applied r.m.s. voltage = 436.84 – 206.74
                                                = 230.1 volt. 

(c) Average power transferred over a complete cycle by the source to inductor is always zero because of phase difference of $\mathrm{\pi }/2$ between voltage and current through the inductor.

(d) Average power transferred over a complete cycle by the source to the capacitor is also zero because of phase difference of $\mathrm{\pi }/2$ between voltage and current through capacitor. 

(e) Total average power absorbed by the circuit is also, therefore zero.

Here given,
Capacitance, $$\begin{gathered} \mathrm{C} = 100 \mathrm{\mu F} = 100 \times {10}^{-6}\mathrm{F} \\ = {10}^{-4}\mathrm{F} \end{gathered}$$ 
Resistance, R = 40 Ω 
 
$$\begin{gathered} \mathrm{Rms} \mathrm{voltage}, {\mathrm{E}}_{\mathrm{v}} = 110 \mathrm{volt} \\ \mathrm{Peak} \mathrm{voltage}, {\mathrm{E}}_0 = \sqrt{2}. {\mathrm{E}}_{\mathrm{v}} = \sqrt{2} \times 110 \mathrm{V} \\ \mathrm{Frequency} \mathrm{of} \mathrm{Ac} \mathrm{supply}, \mathrm{v} = 60 \mathrm{Hz}., \\ \therefore \mathrm{\omega } = 2\mathrm{\pi v} = 120\mathrm{\pi } \mathrm{rad}/\mathrm{s} \\ \mathrm{Peak} \mathrm{current}, {\mathrm{I}}_0 = ? \end{gathered}$$ 

a.) In RC circuit,  as
$\boxed{\mathrm{Z} = \sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{c}}^2}} = \sqrt{{\mathrm{R}}^2+\frac{1}{{\mathrm{\omega }}^2{\mathrm{C}}^2}}$
Therefore,

        $$\begin{gathered} \boxed{{\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\sqrt{{\mathrm{R}}^2+{\displaystyle \frac{1}{{\mathrm{\omega }}^2{\mathrm{C}}^2}}}}} \\ = \frac{\sqrt{2} \times 110}{\sqrt{1600+{\displaystyle \frac{1}{{\left(120 \mathrm{\pi } \times {10}^{-4}\right)}^2}}}} \end{gathered}$$
       ${\mathrm{I}}_0 = 3.24 \mathrm{amp}.$ 

b.) In RC circuit, voltage lags behind the current by phase angle ϕ,
where ϕ is given by,

$$\begin{gathered} \boxed{\tan \mathrm{\varphi } = \frac{1/\mathrm{\omega C}}{\mathrm{R}}} \\ = \frac{1}{\mathrm{\omega CR}} \\ = \frac{1}{120 \mathrm{\pi } \times {10}^{-4} \times 40} \\ = 0.6628 \end{gathered}$$ 

$$\begin{gathered} \Rightarrow \mathrm{\varphi } = {\tan }^{-1}(0.6628) = 33.5 ° \\ = \frac{33.5 \mathrm{\pi }}{180}\mathrm{rad}. \end{gathered}$$ 
Hence, 

$$\begin{gathered} \boxed{\mathrm{Time} \mathrm{lag} = \frac{\mathrm{\varphi }}{\mathrm{\omega }}} \\ = \frac{33.5 \mathrm{\pi }}{180 \times 120 \mathrm{\pi }} \\ = 1.55 \times {10}^{-3} \sec . \end{gathered}$$

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.

i.e., we can see that, ϕ is nearly zero at high frequency.

It is clear from here that at high frequency, capacitor acts like a conductor.
For a D.C. circuit, after steady state has been reached, ω = 0.
Hence, ${\mathrm{\chi }}_{\mathrm{C}} = \frac{1}{\mathrm{\omega C}} = \infty$
Therefore, capacitor C amounts to an open circuit.