 Physics POWER IN AC CIRCUIT: THE POWER FACTOR, LC OSCILLATIONS, TRANSFORMERS FOR CBSE-NCERT

### Topic covered

color{blue}{star} POWER IN AC CIRCUIT: THE POWER FACTOR
color{blue}{star} LC OSCILLATIONS
color{blue}{star} TRANSFORMERS

### POWER IN AC CIRCUIT: THE POWER FACTOR

color{blue} ✍️We have seen that a voltage v = v_m sin omegat applied to a series RLC circuit drives a current in the circuit given by i= i_m sin(omegat + phi) where

i_m = (v_m)Z and phi = tan^(-1) \ (X_C-X_L)/R

color {blue}{➢➢}Therefore, the instantaneous power p supplied by the source is

p = vi = (v_m sin omegat)xx [i_m sin(omegat+phi)]

color {blue}{= (v_mi_m)/2 [cosphi - cos(2omegat+phi)]}

............(7.37)

color{blue} ✍️The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.37). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore,

 P = (v_m t_m)/2 cos phi = v_m/sqrt 2 t_m/sqrt 2 cos phi

color {blue}{= VI cos phi}

.................[7.38(a) ]

color {blue}{➢➢}This can also be written as,

color {blue}{P= I^2 cosphi}

...............[7.38(b)]

color {blue}{➢➢}So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle f between them. The quantity cosphi is called the power factor. Let us discuss the following cases:

color {brown}{"Case (i)"} Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case phi = 0, cos phi= 1. There is maximum power dissipation.

color {brown}{"Case (ii)"} Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is pi//2. Therefore, cos f = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current.

color {brown}{"Case (iii)"} LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.38) where phi = tan–1 (X_c – X_L )//R. So, f may be non-zero in a RL or RC or  RCL circuit. Even in such cases, power is dissipated only in the resistor.

color {brown}{"Case (iv)"} Power dissipated at resonance in LCR circuit: At resonance X_c – X_L= 0, and phi = 0. Therefore, cos phi = 1 and P = I^2Z = I^2 R. That is, maximum power is dissipated in a circuit (through R) at resonance.
Q 3231391222 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.

(b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. Solution:

(a) We know that P = I V cos phi where cos phi is the power factor.

To supply a given power at a given voltage, if cosf is small, we have to increase current accordingly. But this will lead to large power loss (I^2R) in transmission.

(b) Suppose in a circuit, current I lags the voltage by an angle phi. Then power factor cos phi =R//Z.
We can improve the power factor (tending to 1) by making Z tend to R.

Let us understand, with the help of a phasor diagram (Fig. 7.17) how this can be achieved.

Let us resolve I into two components. I_p along the applied voltage V and I_q perpendicular to the applied voltage. I_q as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I'_q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I'_q cancel each other and P is effectively I_p V
Q 3211391229 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Omega, L = 25.48 mH, and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and (d) the power factor. Solution:

To find the impedance of the circuit, we first calculate X_L and X_C.

X_L= 2 pi ne L

=2 xx 3.14 xx 50 xx 25.48 xx 10^(-3) Omega = 8 Omega

X_C = 1/(2 pi nu C)

=1/( 2 xx 3.14 xx 50 xx 796 xx 10^(-6))= 4 Omega

Therefore,

Z= sqrt( R^2 +(X_L - X_C)^2)= sqrt(3^2 +(8 -4)^2)

= 5 Omega

(b) Phase difference, phi = tan^(–1) \ (X_C- X_L)/R

= tan^(-1) ((4-8)/3)=-53.1^o

Since phi is negative, the current in the circuit lags the voltage across the source.

(c) The power dissipated in the circuit is  P = I^2 R

Now, I= (i_m)/(sqrt 2) = 1/(sqrt 2) (283/5) = 40 A

Therefore,  P = (40A )^2 xx 3Omega = 4800 W

(d) Power factor = cos phi = cos 53.1° = 0.6
Q 3231491322 Suppose the frequency of the source in the previous example can be varied.

(a) What is the frequency of the source at which resonance occurs?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution:

(a) The frequency at which the resonance occurs is

omega_0 = 1/(sqrt(LC)) = 1/(sqrt (25.48 xx 10^(-3) xx 796 xx 10^(-6))

= 222.1 rad//s

nu_r =(omega_0)/(2 pi) =(221.1)/( 2 xx 3.14) Hz = 35.4 Hz

(b) The impedance Z at resonant condition is equal to the resistance:

Z = R = 3 Omega

The rms current at resonance is

=V/Z =V/R =(283/(sqrt 2)) 1/3 =66.7 A

The power dissipated at resonance is  P=I^2 xx R = (66.7) ^2 xx 3 = 13.35 kW

You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7.8.
Q 3251491324 At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution:

The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

### LC OSCILLATIONS color{blue} ✍️We know that a capacitor and an inductor can store electrical and magnetic energy, respectively.

color{blue} ✍️When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems.

color{blue} ✍️Let a capacitor be charged q_m (at" "t = 0) and connected to an inductor as shown in Fig. 7.18. The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to current in the circuit. Let q and i be the charge and current in the circuit at time t. Since di//dt is positive, the induced emf in L will have polarity as shown, i.e., vb < va. According to Kirchhoff’s loop rule,

color {blue}{q/C - L (di)/(dt) =0}

..............(7.39)

i = – (dq//dt) in the present case (as q decreases, i increases). Therefore, Eq. (7.39) becomes:

color {blue}{(d^2q)/(dt^2) + 1/(LC) q=0}

............(7.40)

color {blue}{➢➢}This equation has the form (d^2x)/(dt^2) + omega_(0)^(2) x=0 for a simple harmonic oscillator. The charge, therefore, oscillates with a natural frequency

color {blue}{omega_0 = 1/(LC)}

..............(7.41) and varies sinusoidally with time as

color {blue}{q = q_m cos(omega_0t+phi)}

..........(7.42)

color {blue}{➢➢}where q_m is the maximum value of q and phi is a phase constant. Since q = q_m "at" " " t = 0, we have cos phi =1 or phi = 0. Therefore, in the present case

color {blue}{q = q_m cos(omega_0t)}

............(7.43)

The current i (=-(dq)/(dt)) is given by

color {blue}{i = i_m sin (omega_0t)}

.............(7.44)

Where i_m sin (omega_0t)

color {blue}{➢➢}Let us now try to visualise how this oscillation takes place in the circuit.

color {blue}{➢➢}Figure 7.19(a) shows a capacitor with initial charge q_m connected to an ideal inductor. The electrical energy stored in the charged capacitor is U_E = 1/2 (q_(m)^(2))/C Since, there is no current in the circuit, energy in the inductor is zero. Thus, the total energy of LC circuit is

U = U_E = 1/2 (q_(m)^(2))/C color {blue}{➢➢}At t = 0, the switch is closed and the capacitor starts to discharge [Fig. 7.19(b)].

color {blue}{➢➢}As the current increases, it sets up a magnetic field in the inductor and thereby, some energy gets stored in the inductor in the form of magnetic energy : U_B = (1/2) Li^2.

color {blue}{➢➢}As the current reaches its maximum value i_m, (at" "t = T//4) as in Fig. 7.19(c), all the energy is stored in the magnetic field: U_B = (1//2) Li^2 m. You can easily check that the maximum electrical energy equals the maximum magnetic energy.
The capacitor now has no charge and hence no energy. The current now starts charging the capacitor, as in Fig. 7.19(d). This process continues till the capacitor is fully charged (at t = T//2) [Fig. 7.19(e)].

color {blue}{➢➢}But it is charged with a polarity opposite to its initial state in Fig. 7.19(a). The whole process just described will now repeat itself till the system reverts to its original state. Thus, the energy in the system oscillates between the capacitor and the inductor

color{blue} ✍️The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. The lower part of each figure in Fig. 7.19 depicts the corresponding stage of a mechanical system (a block attached to a spring). As noted earlier, for a block of a mass m oscillating with frequency omega_0, the equation is (d^2x)/(dt^2) + omega_(0)^(2)x=0

color {blue}{➢➢}Here omega_0 = sqrt(K//m) and k is the spring constant. So, x corresponds to q. In case of a mechanical system F = ma = m (dv//dt) = m (d^2x//dt^2).

color {blue}{➢➢}For an electrical system, e = –L (di//dt ) = –L (d2q//dt 2). Comparing these two equations, we see that L is analogous to mass m: L is a measure of resistance to change in current. In case of LC circuit, omega_0 = 1//sqrt(LC) and for mass on a spring omega_0 = sqrt(k//m) So, 1//C is analogous to k.

color{blue} ✍️The constant k (=F//x) tells us the (external) force required to produce a unit displacement whereas 1//(=V//q) tells us the potential difference required to store a unit charge. Table 7.1 gives the analogy between mechanical and electrical quantities. color{brown} {"Note"} that the above discussion of LC oscillations is not realistic for two reasons :

color{blue} {(i)} Every inductor has some resistance. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the oscillations finally die away.

color{blue} {(ii)} Even if the resistance were zero, the total energy of the system would not remain constant. It is radiated away from the system in the form of electromagnetic waves (discussed in the next chapter). In fact, radio and TV transmitters depend on this radiation. ### TRANSFORMERS

color{blue} ✍️For many purposes, it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value. This is done with a device called "transformer" using the principle of mutual induction.

color{blue} ✍️A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.20(a) or on separate limbs of the core as in Fig. 7.20(b).
One of the coils called the primary coil has N_p turns. The other coil is called the secondary coil; it has N_s turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. color{blue} ✍️When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it.

color{blue} ✍️The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings.
Let phi be the flux in each turn in the core at time t due to current in the primary when a voltage v_p is applied to it. Then the induced emf or voltage es, in the secondary with N_s turns is

color {blue}{epsilon_s = - N_s (dphi)/(dt)}

.............(7.45)

color {blue}{➢➢}The alternating flux phi also induces an emf, called back emf in the primary. This is

color {blue}{epsilon_s = - N_p (dphi)/(dt)}

.............(7.46)

color {blue}{➢➢}But epsilon_p = v_p. If this were not so, the primary current would be infinite since the primary has zero resistance(as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation epsilon_s = v_s

color {blue}{➢➢}where vs is the voltage across the secondary. Therefore, Eqs. (7.45) and (7.46) can be written as

color {blue}{v_s = - N_s(dphi)/(dt)}

..............[7.45(a)]

color {blue}{v_p = - N_p(dphi)/(dt)}

................[7.46(a)]

color {blue}{➢➢}From Eqs. [7.45 (a)] and [7.46 (a)], we have

color {blue}{(v_s)/(v-p) = (N_s)/(N_P)}

...............(7.47)

color{brown}bbul {"Note"} that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v,

color {blue}{ i_pv_p = i_sv_s}

.............(7.48)

color {blue}{➢➢}Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (7.47) and (7.48), we have

color {blue}{(i_P)/(i_S) = (V_s)/(V_P)= (N_S)/(N_P)}

..........(7.49)

color {blue}{➢➢}Since i and v both oscillate with the same frequency as the ac source, Eq. (7.49) also gives the ratio of the amplitudes or rms values of corresponding quantities.

color {blue}{➢➢}Now, we can see how a transformer affects the voltage and current. We have:

color {blue}{V_S = (N_S)/(N_P) V_P and I_S = (N_P)/(N_S) I_P}

................(7.50)

color {blue}{➢➢}That is, if the secondary coil has a greater number of turns than the primary (N_s > N_p), the voltage is stepped up (V_s > V_p).

This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (N_p//N_s < 1 and I_s < I_p).

color {blue}{➢➢}For example, if the primary coil of a transformer has 100 turns and the secondary has 200 turns, N_s//N_p = 2 and N_p//N_s=1//2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A.

color{blue} ✍️If the secondary coil has less turns than the primary (N_s < N_p), we have a step-down transformer. In this case, V_s < V_p and I_s > I_p. That is, the voltage is stepped down, or reduced, and the current is increased.

color{blue} ✍️The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons:

color{blue} {(i) bbul{"Flux Leakage:"}} There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other.

color{blue} {(ii) bbul{"Resistance of the windings:"}} The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I^2R). In high current, low voltage windings, these are minimised by using thick wire.

color{blue} {(iii) bbul{"Eddy currents"}} The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a laminated core.

color{blue} {(iv) bbul{"Hysteresis"}} The magnetisation of the core is repeatedly reversed by the alternating magnetic field.
The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss.

color{blue} ✍️The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I^2R loss is cut down).
It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. 