Topic covered

`color{blue}{star}` RESONANCE
`color{blue}{star}` SHARPNESS OF RESONANCE


`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)`


`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))`


`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)`


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

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


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

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


`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])`


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


`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))`


`color {blue}{➢➢}`where

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


`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`


`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)`


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))`


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



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


`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)`


`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)`


`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)`


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

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


`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)`


`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))`


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


`color {blue}{➢➢}`where

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


`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.


`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)`


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


`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.


`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_1`and `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{blue} ✍️`We can approximate `color{green}((1+(Deltaomega)/(omega_0))^(-1) as (1 - (Deltaomega)/(omega_0))` since `color{green}((Deltaomega)/(omega_0) `<< `1)`. Therefore


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

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


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

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


`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)`


`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

`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.