Question

A series LCR circuit with $L=012H$, $C=480nF$, $R=23\Omega$ is connected to a $230V$ variable frequency supply.

(a) What is the source frequency for which the current amplitude will be maximum? Also, obtain this maximum value.

(b) What is the source frequency for which average power absorbed by the circuit will be maximum? Obtain the value of this maximum power.

(c) For which frequencies of the source, the power transferred to the circuit will be half of the power at resonant frequency? What is the current amplitude at these frequencies?

(d) What is the Q-factor of the given circuit?

Answer 1

$a)L=012H,C=480nF,=480\times {10}^{-9},R=3vr=230v$

$V_0=\sqrt{2}\times 230=325.22V$

Current blowing in the circuit $I_0=\frac{V_0}{\sqrt{2^2+{(wl-\frac{1}{wl})}^2}}$

At reference $w_r=\frac{1}{w_r}$

$\therefore w_r=\frac{1}{\sqrt{Lc}}=\frac{1}{\sqrt{0.12}\times 480\times {10}^{-9}}=4166.67$

$V_R=\frac{wr}{2\pi }=\frac{4166.67}{2\times 3.14}=663.46$

Maximum current $I_{0mx}=\frac{V_o}{R}=\frac{325.22}{23}=14.14A$

$b)$max average power observed The circuit

$(P_{av}{)}_{max}=\frac{1}{2}(Io{)}_{max}^R$

$=\frac{1}{2}(14.14{)}^2\times 23=2299.3w$

$c)$power translated to the circuit n half the power at resonant

$w_R\pm △w=2\pi (v_R\pm △v)$ $△w=R/2L$

= $\frac{23}{2\times 0.12}=95.83$

chance in frequency

$△v=\frac{1}{2\pi }△w=\frac{95.83}{2\pi }=15.26$

$v_R+△v=663.48+15.26=678.74$

$v_R-△v=663.48-15-26=648-22$

At the frequency current $ampw$given as

$I^{\prime }=\frac{1}{\sqrt{2}}\left(I_0\right)=\frac{14.14}{1.414}=10A$

$d)Q$ factor given as

$Q=\frac{w_{r}L}{R}$

$\frac{4166.67\times 0.12}{23}=21.74$

$\therefore factorr=21.74$