Physics AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT, RESONANCE, SHARPNESS OF RESONANCE FOR CBSE-NCERT

### Topic covered

color{blue}{star} AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
color{blue}{star} RESONANCE
color{blue}{star} SHARPNESS OF RESONANCE

### AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT

color{blue} ✍️Figure 7.12 shows a series LCR circuit connected to an ac source e. As usual, we take the voltage of the source to be v = v_m sin omega t.

color {blue}{➢➢}If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule:

color{green}(L (di)/(dt) +i R + q/C =v)

.............(7.20)

color{blue} ✍️We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this problem by two methods. First, we use the technique of phasors and in the second method, we solve Eq. (7.20) analytically to obtain the time– dependence of i .

color{green}bbul("Phasor-diagram solution")

color {blue}{➢➢}From the circuit shown in Fig. 7.12, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be

color{green}(i = i_m sin(omegat + phi))

............(7.21)

color {blue}{➢➢}where phi is the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case.

color{blue} ✍️Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let color{green}(V_L, V_R, V_C,) and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that V_R is parallel to color{green}(I, V_C) is color{green}(pi//2) behind I and color{green}(V_L) is color{green}(pi//2) ahead of color{green}(I. V_L, V_R, V_C) and I are shown in Fig. 7.13(a) with appropriate phase relations.

color {blue}{➢➢}The length of these phasors or the amplitude of color{green}(V_R V_C) and V_L are;

color{green}(v_(Rm) = i_m R, v_(Cm) = i_m X_C v_(Lm) = i_m X_L)

..........(7.22)

color {blue}{➢➢}The voltage Equation (7.20) for the circuit can be written as

color{green}(v_L +v_R v_C =v)

..............(7.23)

color {blue}{➢➢}The phasor relation whose vertical component gives the above equation is

color{green}(V_L +V_R V_C =v)

............(7.24)

color {blue}{➢➢}This relation is represented in Fig. 7.13(b).

color {blue}{➢➢}Since V_C and V_L are always along the same line and in opposite directions, they can be combined into a single phasor color{green}((V_C + V_L)) which has a magnitude color{green}(|v_(Cm) – v_(Lm)|).

color {blue}{➢➢}Since V is represented as the hypotenuse of a right-traingle whose sides are V_R and color{green}((VC + VL)), the pythagorean theorem gives:

color{green}(v_(m)^(2) = v_(Rm)^(2) + (v_(Cm) - v_(Lm)^2)

color {blue}{➢➢}Substituting the values of v_(Rm), v_(Cm,) and v_(Lm) from Eq. (7.22) into the above equation, we have

color{green}(v_(m)^(2) = (i_mR)^2+(i_mX_C-i_mX_L)^2)
color{green}(= i_(m)^(i) [ R^2+(X_C-X_L)^2])

or

color{green}(i_m = (v_m)/(sqrtR^2 + (X_C-X_L)^2))

............[7.25(a)]

color {blue}{➢➢}By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit:

color{green}(i_m = (v_m))

............[7.25(b)]

color {blue}{➢➢}where

color{green}(Z = sqrt(R^2+(X_C-X_L)^2))

...........(7.26)

color {blue}{➢➢}Since phasor I is always parallel to phasor V_R . the phase angle phi is the angle between V_R and V and can be determined from
Fig. 7.14:

 tan phi = (v_(Cm) - v_(Lm))/(v_(Rm)

color {blue}{➢➢} Using Eq. (7.22), we have

 tan phi = (X_C - X_L)/R

.............(7.27)

color {blue}{➢➢}Equations (7.26) and (7.27) are graphically shown in Fig. (7.14). This is called Impedance diagram which is a right-triangle with Z as its hypotenuse.

color {blue}{➢➢}Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified

color {blue}{➤➤}If X_C > X_L , phi is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage.
color {blue}{➤➤}If X_C < X_L , phi is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.

color {blue}{➢➢}Figure 7.15 shows the phasor diagram and variation of v and i with ω t for the case X_C > X_L.

color {blue}{➢➢}Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages.
First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t_1, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors.

color {blue}{➢➢}The solution so obtained is called the steady-state solution. This is not a general solution. Additionally, we do have a transient solution which exists even for v = 0. The general solution is the sum of the transient solution and the steady-state solution.
After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution

color{brown}bbul("Analytical solution")

The voltage equation for the circuit is color{green}(L(di)/(dt) +Ri + q/c =v)

=color{green}(v_m sin omegat)

color {blue}{➢➢}We know that color{green}(i = dq//dt.) Therefore, color{green}((di)/dt = (d^2q)/dt^2). Thus, in terms of q, the voltage equation becomes

color{green}(L(d^2q)/(dt^2) + (R_dp)/(dt) + q/c= v_m sin omegat)

.........(7.28)

This is like the equation for a forced, damped oscillator, [see Eq. {14.37(b)} in Class XI Physics Textbook]. Let us assume a solution

color{green}(q = q_m sin (omegay+theta))

..........[7.29(a)]

color {blue}{➢➢}so that color{green}((dp)/(dt) = q_m omega cos (omegat+theta))

..........[7.29(b)]

and

color{green}((d^2q)/(dt^2) = - q_m omega^2 sin (omegat+theta))

..........[7.29(c)]

color {blue}{➢➢}Substituting these values in Eq. (7.28), we get

color{green}(q_m omega[R cos (omegat+theta) + (X_C - X_L ) sin (omegat+theta) = v_m sin omega t)

...........(7.30)

color {blue}{➢➢}where we have used the relation color{green}(X_c= 1//omegaC, X_L = omega L.) Multiplying and

dividing Eq. (7.30) by color{green}(Z = sqrt(R^2+(X_C-X_L)^2) we have

color{green}(q_m omegaZ [R/Z cos (omegat+theta) + (X_C-X_L)/Z sin (omegat+theta)] = v_m sin omegat)

.........(7.31)

color {blue}{➢➢}Now, let color{green}(R/Z = cos phi)

and color{green}((X_C-X_L)/Z = sin phi)

color{green}(phi = tan^(-1) " " (X_C-X_L)/R)

...........(7.32)

color {blue}{➢➢}Substituting this in Eq. (7.31) and simplifying, we get:

color{green}(q_m omegaZcos (omegat+theta-phi) = v_m sin omegat)

............(7.33)

color {blue}{➢➢}Comparing the two sides of this equation, we see that

color{green}(v_m = q_m omegaZ= i_m Z)

color {blue}{➢➢}where color{green}(i_m = q_m omega).........[7.33(a)]

color {blue}{➢➢}and

color{green}(theta phi = pi /2 or theta =- pi/2 + phi)

..........[7.33(b)]

color {blue}{➢➢}Therefore, the current in the circuit is

color{green}(i = (dq)/(dt) = q_m cos (omegat + theta))

color{green}(= i_m cos(omegat+theta))

or

color{green}(i = i_m sin (omega t+phi))

..........(7.34)

color {blue}{➢➢}where

color{green}(i_m = (v_m)/Z = (v_m)/(sqrt(R^2+(X_C-X_L)^2))

.............[7.34(a)]

color {blue}{➢➢}and color{green}(phi = tan^(-1) ((X_X-X_L)/(R)))

color {blue}{➢➢}Thus, the analytical solution for the amplitude and phase of the current in the circuit agrees with that obtained by the technique of phasors.

### RESONANCE

color{blue} ✍️An interesting characteristic of the series RLC circuit is the phenomenon of resonance.
The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the system’s natural frequency.

color{blue} ✍️If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large. A familiar example of this phenomenon is a child on a swing.

color{blue} ✍️The swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging, the amplitude of the swinging will be large.
For an RLC circuit driven with voltage of amplitude v_m and frequency w, we found that the current amplitude is given by

color{green}(i_m = (v_m)/Z = (v_m)/(sqrt(R^2+(X_C-X_L)^2))

with color{green}(X_c = 1/omega C and XL = omega L). So if w is varied, then at a particular frequency

color{green}(omega_0 X_C = X_L) and the impedance is minimum color{green}((Z = sqrt(R^2+0^2) = R)) frequency is called the resonant frequency:

color{green}(X_C = X_L or 1/(omega_C) = omega_0 L)

or

color{green}(omega_0 = 1/(sqrt (LC)))

............(7.35)

color {blue}{➢➢}At resonant frequency, the current amplitude is maximum; color{green}(im = v_m//R.)

color {blue}{➢➢}Figure 7.16 shows the variation of i_m with omega in a RLC series circuit with color{green}(L = 1.00 mH, C = 1.00 nF) for two values of color{green}(R: (i) R = 100 Omega) and color{green}((ii) R = 200 Omega). For the source applied color{green}(v_m =)

color{green}(100 V. omega_0) for this case is color{green}(1/(sqrt(LC))= 1.00xx10^6 rad//s)

color{blue} ✍️ We see that the current amplitude is maximum at the resonant frequency. Since color{green}(i_m = v_m // R) at resonance, the current amplitude for case (i) is twice to that for case (ii).

color{blue} ✍️Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations.

color{blue} ✍️The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received.

color {blue}{➢➢}When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum.

color{brown}bbul{"Note :"}
color{blue} ✍️It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is v_m//R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.

### SHARPNESS OF RESONANCE

color{blue} ✍️The amplitude of the current in the series LCR circuit is given by

color{green}(i_m = (v_m)/(sqrt(R^2 + (omegaL- 1/(omegac))^2)

color {blue}{➢➢}and is maximum when color{green}(omega = omega_0 = 1//sqrt(LC)) The maximum value is color{green}(i_(m)^(max) = v_m //R)

color {blue}{➢➢}For values of w other than omega_0, the amplitude of the current is less than the maximum value. Suppose we choose a value of omega for which the current amplitude is 1// sqrt2 times its maximum value.
At this value, the power dissipated by the circuit becomes half. From the curve in Fig. (7.16), we see that there are two such values of w, say, omega_1and omega_2, one greater and the other smaller than omega_0 and symmetrical about omega_0. We may write

color{green}(omega_1 = omega_0 + Deltaomega)
color{green}(omega_2 = omega_0 + Deltaomega)

color {blue}{➢➢}The difference color{green}(omega_1 – omega_2 = 2Deltaomega) is often called the bandwidth of the circuit.
color {blue}{➢➢}The quantity color{green}((omega_0 // 2Deltaomega)) is regarded as a measure of the sharpness of resonance. The smaller the

color{green}(Deltaomega,) the sharper or narrower is the resonance.

color{blue} ✍️To get an expression for Delta omega, we note that the current amplitude i_m is

color{green}((1//sqrt2) i_(m)^(max) omega_1= omega_0 + Deltaomega) Therefore,

at color{green}(omega_1. i_m = (v_m)/(sqrt(R^2+(omega_1L- 1/(omega_1C))^2)

color{green}(= i_(m)^(max)/(sqrt2)= (v_m)/(Rsqrt2)

or color{green}(sqrt(R^2+(omega_1L- 1/(omega_1C))^2)=Rsqrt2)

or color{green}(R^2+(omega_1L- 1/(omega_1C))^2= 2R^2)

color{green}(omega_1L- 1/(omega_1C)=R) which may be written as,

color{green}((omega_0+Deltaomega)L - 1/((omega_0+Deltaomega)C)= R)

color{green}(omega_0L (1 +(Deltaomega)/(omega_0)) - 1/(omega_0C(1+(Deltaomega)/(omega_0)))=R)

color {blue}{➢➢}Using color{green}(omega_(0)^(2) = 1/(LC)) in the second term on the left hand side, we get

color{green}(omega_0L(1+(Deltaomega)/(omega_0))-(omega_0L)/(1+(Deltaomega)/(omega_0))=R)

color{blue} ✍️We can approximate color{green}((1+(Deltaomega)/(omega_0))^(-1) as (1 - (Deltaomega)/(omega_0)) since color{green}((Deltaomega)/(omega_0) << 1). Therefore

color{green}(omega_0L(1+(Deltaomega)/(omega_0))-omega_0L(1-(Deltaomega)/(omega_0))R)

or color{green}(omega_0 L(2Deltaomega)/(omega_0) =R)

color{green}(Deltaomega= R/(2L))

............[7.36(a)]

color{blue} ✍️The sharpness of resonance is given by,

color{green}((omega_0)/(2Deltaomega) = (omega_0L)/R)

...........[7.36(b)]

color {blue}{➢➢}The ratio color{green}((omega_0L)/R) is also called the quality factor, Q of the circuit.

color{green}(Q = (omega_0L)/R)

............[7.36(c)]

color {blue}{➢➢}From Eqs. [7.36 (b)] and [7.36 (c)], we see that color{green}(2 Delta omega= (omega_0)/Q) So, larger the value of Q, the smaller is the value of color{green}(2 Delta omega) or the bandwidth and sharper is the resonance. Using 2 omega_(0)^(2) = 1//L C , Eq. [7.36(c)] can be equivalently expressed as color{green}(Q = 1/omega_0CR).

color{blue} ✍️We see from Fig. 7.15, that if the resonance is less sharp, not only is the maximum current less, the circuit is close to resonance for a larger range color{green}(Deltaomega) of frequencies and the tuning of the circuit will not be good.

color {blue}{➢➢}So, less sharp the resonance, less is the selectivity of the circuit or vice versa. From Eq. (7.36), we see that if quality factor is large, i.e., R is low or L is large, the circuit is more selective.
Q 3178256106

A resistor of 200 Omega and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.
Class 12 Chapter 7 Example 6
Solution:

Given
R = 2000 Omega C = 15muF = 15.0 xx 10^(-5)F

V = 220 V, n = 50Hz

(a) In order to calculate the current, we need the impedance of the circuit. It is

Z= sqrt(R^+C_(C)^(2)) = sqrt(R^2+(2pirvC)^(-2))

= sqrt((200Omega)^2+(2x3.14xx50xx10^(-6)F)^2)

= 291.5Omega

Therefore, the current in the circuit is

I= V/Z (220V)/(291.5Omega) = 0.755A

(b) Since the current is the same throughout the circuit, we have

V_R = IR = (0.755A)(200Omega)=0.755A

V_R = IR = (0.755A)(212.3Omega) = 160.3C

The algebraic sum of the two voltages, V_R and V_C is 311.3 V which is
more than the source voltage of 220 V. How to resolve this paradox?
As you have learnt in the text, the two voltages are not in the same
phase. Therefore, they cannot be added like ordinary numbers. The
two voltages are out of phase by ninety degrees. Therefore, the total
of these voltages must be obtained using the Pythagorean theorem:

V_(R+C) = sqrt(V_(R)^(2) +V_(C)^(2)
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.