Question

For an A.C. circuit containing an inductance L, a capacitance C and a resistance R connected in series, establish the formula for impedance of the circuit and write the relationship between the alternating e.m.f. and currents in each of the following case when :
(i) $wL>\frac{1}{\omega C}$,
(ii) $wL<\frac{1}{\omega C}$,
(iii) $wL=\frac{1}{\omega C}$.

Answer 1

Resonant circuit: In L-C-R circuit, when the value of impidence is least, then the current becomes maximum this phenomenon is called resonance.

In this state the frequency of the applied E.M.F. is equal to the natural frequency of the L-C circuit called resonante frequency and this circuit is called resonant electric circuit.

In the figure an inductance L, a capacitance C and a resistance R are connected in series with an alternating voltage.

At any instant alternating voltage is given as

$V=V_{o}sin\omega t$ .........(1)

Where $V_o$ is the peak value of alternating voltage at any instant, if the current in the circuit be I, then. The potential difference across the inductance L is.

$V_L=X_{L}I$ ........(2)

Similarly potential difference across the capacitance C is $V_C$

$V_C=X_{C}I$ ...........(3)

and the potential difference across the resistance R is $V_R$

$V_R=RI$ .......(4)

Now $V_R$ and I are in the same phase while $V_L$ leading ahead I by ${90}^o$ and $V_C$ logging behind I by ${90}^o$ hence the angle between $V_L$ and VC is ${180}^o$ i.e., $V_L$ and $V_C$ are in opposite phase. Now the resultant of $V_L$ and $V_C$ is $V_L-V_C$ then the angle between $V_L-V_C$ and $V_R$ is ${90}^o$.

If the resultant of $(V_L-V_C)$ and $V_R$ is V.

Then according to the figure (2) by pythagoras theorem.

$V^2=(V_L-V_C{)}^2+V_R^2$ .......(5)

from equation (2), (3), (4) and (5)

$V^2=(X_{L}I-X_{C}I{)}^2+(RI{)}^2$

$V^2=I^2(X_L-X_C{)}^2+I^{2}R$

$\frac{V^2}{I^2}=(X_L-X_C)2+R2$

$\frac{V}{I}=\sqrt{(X_L-X_C{)}^2+R^2}$ ..........(6)

On comparing equation (6) with ohms law, the right hand side of equation (6) shows that the effective resistance of this circuit which is called impedance of this circuit and denoted by $Z_{L-C-R}$

then, $Z_{L-C-R}=\sqrt{(X_L-X_C{)}^2+R^2}$

but, $X_L=WL$ and $X_C=\frac{1}{{\omega }_C}$

$Z_{L-C-R}=\sqrt{{(WL-\frac{1}{\omega c})}^2+R^2}$

equation (7) and (8) are the required expression for the impidence of the this circuit.

Phase difference between V and I-According to the above the circuit from (2) and current I in the leggs behind the potential difference V by phase angle $\Phi$ then figure.

$tan\Phi =\frac{V_L-V_C}{V_R}=P$

$\Phi =ta{n}^{-1}\left(\frac{V_L-V_C}{V_R}\right)$

From equation (2), (3), (4) and (5)

$\Phi =ta{n}^{-1}\left(\frac{X_{L}I-XC.I}{RI}\right)$

$\Phi =ta{n}^{-1}\left(\frac{X_L-X_C}{RI}\right)$

but $X_L=WL,X_C=\frac{1}{WC}$

then,

$\Phi =ta{n}^{-1}\left(\frac{WL-\frac{1}{WC}}{R}\right)$

equation (9), (10), (11) are required expression for phase difference for current. Thus in the L-C-R circuit the current can be expressed by the following equation.

$I=I_{o}sin(wt-\Phi )$

Where $\Phi =ta{n}^{-1}\left(\frac{V_L-V_C}{V_R}\right)$

and $I_o=\frac{V_o}{Z_{L-C-R}}$

here

$Z_{L-R-C}=\sqrt{{(wL-\frac{1}{wC})}^2+R^2}$

Resonance frequency: For resonance condition inductive reactance $X_L$ and capacitative reactance X equal in this circuit so the resistance of this circuit will be least then the current becomes minimum i.e., in resonance condition.

$X_L=X_C$

but $X_L=w.L$

$X_C=\frac{1}{wC}$

$w^2=\frac{1}{LC}$

$w=\frac{1}{\sqrt{LC}}$

but $2\pi f=w$

$2\pi f=\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{L.C}}$

(where f is resonance frequency of this circuit)

This is required expression for resonant frequency.