Question

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

Answer 1

In this type of questions we use the following formula.
Impedance of series L-C-R circuit is below.
$Z = \sqrt {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}}$
Here $\omega$ is frequency.

Complete step by step solution:
Let voltage across the inductor, capacitor and resistor is ${V_L},{V_{C,}}{V_R}$ respectively. Current will be the same in each component because of series connection. The voltage across an ideal inductor (L) is $90^o$ ahead of an AC current through it and voltage across an ideal capacitor (C) is $90^o$ behind AC current through it. Hence, ${V_L}$ and ${V_C}$ is 180 degrees from each other. As a result their resultant will be at ${90^ \circ }$ with${V_R}$.
At any instant let alternating voltage is given as follows.
$V = {V_0}\sin \omega t$
So, we can write the following.
${V^2} = V_R^2 + {({V_c} - {V_L})^2}$
$V = \sqrt {V_R^2 + {{({V_c} - {V_L})}^2}}$ …………..(1)
We know the following also.
${V_R} = RI$, ${V_c} = {X_C}I$ and ${V_L} = {X_L}I$ also, ${X_C} = \dfrac{1}{{\omega C}}$ and ${X_L} = \omega L$
Now let us substitute the values in (1).
$V = \sqrt {{{(RI)}^2} + {{({X_C}I - {X_L}I)}^2}}$
Now, impedance is given by the following formula.
$Z = \dfrac{V}{I} = \sqrt {{{(R)}^2} + {{({X_C} - {X_L})}^2}}$
$Z = \sqrt {{R^2} + {{(\dfrac{1}{{\omega C}} - \omega L)}^2}}$ ……………….(2)
The impedance of the L-C-R circuit in series is given by (2).
The phase difference between V and I given below.
$tan\phi = \dfrac{{{V_L} - {V_C}}}{{{V_R}}}$
So the angle can be written as follows.
$\phi = ta{n^{ - 1}}(\dfrac{{{V_L} - {V_C}}}{{{V_R}}})$ ………………….(3)
Now, let us write the phase difference in terms of current I.
$\phi = ta{n^{ - 1}}(\dfrac{{{X_L}I - {X_C}I}}{{RI}})$
$\Rightarrow$ $\phi = ta{n^{ - 1}}(\dfrac{{{X_L} - {X_C}}}{R})$
Now let us substitute the values ${X_C} = \dfrac{1}{{\omega C}}$ and ${X_L} = \omega L$ in equation (4).
We get the following.
$\phi = ta{n^{ - 1}}(\dfrac{{\omega L - \dfrac{1}{{\omega C}}}}{R})$ ……………..(4)
This is the phase difference for current.
Hence, in L-C-R circuit current can be expressed by the following equation.
$I = {I_0}\sin (\omega t - \phi )$
Where, phase difference is given by equation (2) and${I_0} = \dfrac{{{V_0}}}{Z}$ .

(A) Now let us consider the case when $\omega L > \dfrac{1}{{\omega C}}$
In this case the circuit will behave as an inductive circuit. Current flowing through the inductor produces magnetic flux in the opposite direction to the source. When this current becomes large voltage will be (minimum) zero and vice-versa. So voltage leads the current by $90^o$.
We have the phase difference between voltage and current that is given by the following relation. $\phi = ta{n^{ - 1}}(\dfrac{{\omega L - \dfrac{1}{{\omega C}}}}{R})$
In this case current lags $90^o$ behind voltage.

(B) Now let us consider the case when $\omega L < \dfrac{1}{{\omega C}}$
Similarly, in this case the circuit behaves as a capacitive circuit. In this case current leads the voltage by $90^o$.
We have the phase difference between voltage and current is given below.
$\phi = ta{n^{ - 1}}(\dfrac{{\omega L - \dfrac{1}{{\omega C}}}}{R})$.
In this case, voltage lags behind by $90^o$ from current.

(C) Now, let us consider a case when ${X_L} = {X_C}$
Similarly, we have the phase difference between voltage and current is given by. $\phi = ta{n^{ - 1}}(\dfrac{{{X_L} - {X_C}}}{R})$
After applying the condition, we get the following.
$\phi = ta{n^{ - 1}}(\dfrac{0}{R})$
Therefore, current will be in phase with applied voltage.
This is the condition for resonance. Therefore, resistance of the circuit will be least and current becomes large.
Now, let us substitute the values ${X_L} = \omega L$,${X_C} = \dfrac{1}{{\omega C}}$ in ${X_L} = {X_C}$
So, we get the following after solving.
$\omega L = \dfrac{1}{{\omega C}}$ $\Rightarrow$ ${\omega ^2} = \dfrac{1}{{LC}}$
Hence, we get the value of angular frequency as below.
$\omega = \dfrac{1}{{\sqrt {LC} }}$ ; This is the angular frequency.
Now let us find the general frequency using the formula $2\pi f = \omega$
$2\pi f = \dfrac{1}{{\sqrt {LC} }}$ $\Rightarrow$ $f = \dfrac{1}{{2\pi \sqrt {LC} }}$ (6)
This is called the resonance frequency of the circuit.
Hence, phase of current and voltage lies between $-90^o$ to $+90^o$ depending on the values of frequency and other elements.

Note: Voltage ${V_R}$ and current I will be in same phase, voltage ${V_C}$ will lag behind the current by ${90^ \circ }$ and voltage ${V_L}$ will lead the current by ${90^ \circ }$. Don’t get confused while substituting voltages and frequencies in simplifications.