Physics AC VOLTAGE APPLIED TO AN INDUCTOR, AC VOLTAGE APPLIED TO A CAPACITOR FOR CBSE-NCERT - 2

### Topic covered

color{blue}{star} AC VOLTAGE APPLIED TO AN INDUCTOR
color{blue}{star} AC VOLTAGE APPLIED TO A CAPACITOR

### AC VOLTAGE APPLIED TO AN INDUCTOR

color{blue} ✍️ Figure 7.5 shows an ac source connected to an inductor.

color{blue} ✍️Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v = v_m sin omega t. Using the Kirchhoff’s loop rule sum epsilon(t) = 0 and since there is no resistor in the circuit,

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

...........(7.10)

color{blue} ✍️ where the second term is the self-induced Faraday emf in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law (Chapter 6).
Combining Eqs. (7.1) and (7.10), we have

color{blue}{(di)/(dt) = v/L = (v_m)/L sin omega ct}

............(7.11)

color{blue} ✍️ Equation (7.11) implies that the equation for i(t), the current as a function of time, must be such that its slope di//dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by vm//L. To obtain the current, we integrate di//dt with respect to time:

color{purple}{int(di)/(dt) dt = (v_m)/L int sin(omega vt)dt}

color{blue} ✍️ and get,

color{purple}{i = - (u_m)/(omega L) cos (omega t) +} constant

color{blue} ✍️The integration constant has the dimension of current and is time independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. Using

color{purple}{- cos (omega t) = sin (omega t- (pi)/2)} .

color{blue} ✍️ we have

color{blue}{t = t_m= sin (omega t- (pi)/2)}

...........(7.12)

color{blue} ✍️ where i_m= (v_m)/(omega L) is the amplitude of the current. The quantity omega L is analogous to the resistance and is called inductive reactance, denoted by X_L;

color{blue}{X_L = omega L}

............(7.13)

color{blue} ✍️ The amplitude of the current is, then

color{blue}{i_m = (v_m)/(X_L)}

..........(7.14)

color{blue} ✍️The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Omega).
The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current.

color{blue} ✍️A comparison of Eqs. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by p/2 or one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t_1.
The current phasor I is pi//2 behind the voltage phasor V. When rotated with frequency omega counterclockwise, they generate the voltage and current given by Eqs. (7.1) and (7.12), respectively and as shown in Fig. 7.6(b).

color{blue} ✍️ We see that the current reaches its maximum value later than the voltage by one-fourth of a period [T/4= (pi//2)/omega] You have seen that an inductor has reactance that limits current similar to resistance in a dc circuit.
Does it also consume power like a resistance? Let us try to find out.
The instantaneous power supplied to the inductor is

color{purple}{P_L = iv= i_m sin(omegat- pi/2)xxv_m sin(omegat)}

color{purple}{= (i_mv_m)/2 sin (2omega)}

color{blue} ✍️ So, the average power over a complete cycle is

color{purple}{P_L (-(i_mv_m)/2 sin(2omegat))}

color{purple}{= - (i_mv_m)/2 (sin (2omega t))}

color{blue} ✍️ since the average of sin (2 omega t) over a complete cycle is zero.
Thus, the average power supplied to an inductor over one complete cycle is zero.
Figure 7.7 explains it in details.

Q 3148845703

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Class 12 Chapter 7 Example 2
Solution:

The inductive reactance,

X_L = 2pivL=2xx3.14 xx 50 xx25 xx 10^(-3)W

= 7.85W
The rms current in the circuit is

I = V/X_L = (220V)/(7,85Omega)=28A

### AC VOLTAGE APPLIED TO A CAPACITOR

color{blue} ✍️ Figure 7.8 shows an ac source e generating ac voltage v = v_m sin omega t connected to a capacitor only, a purely capacitive ac circuit.

color{blue} ✍️When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current.
That is, a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, as in Fig. 7.8, it limits or regulates the current, but does not completely prevent the flow of charge.

color{blue} ✍️The capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is

color{blue}{v = q/c}

..............(7.15)

color{blue} ✍️ From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal,

color{purple}{v_m sin omega t = q/c}

color{blue} ✍️ To find the current, we use the relatio t = (dq)/(dt)

color{purple}{i = d/(dt) (v_m C sin omega t) = omegaCv_m cos(omega t)}

color{blue} ✍️ Using the relation, cos (omega t) = sin (omega+ pi/2) we have

color{blue}{t = t_m sin (omega+ pi/2)}

...........(7.16)

color{blue} ✍️ where the amplitude of the oscillating current is i_m = omega Cv_m. We can rewrite it as

color{purple}{i_m = (v_m)/(1//omegaC)}

color{blue} ✍️ Comparing it to i_m= v_m//R for a purely resistive circuit, we find that (1//omega C) plays the role of resistance. It is called capacitive reactance and is denoted by X_c,

color{blue}{X_c= 1 // omegaC}

.............(7.17)

color{blue} ✍️ so that the amplitude of the current is

color{blue}{i_m = (v_m)/(X_c)}

..........(7.18)

color{blue} ✍️ The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Omega).
The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance.

color{blue} ✍️A comparison of Eq. (7.16) with the equation of source voltage, Eq. (7.1) shows that the current is pi//2 ahead of voltage. Figure 7.9(a) shows the phasor diagram at an instant t_1. Here the current phasor I is pi//2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.9(b) shows the variation of voltage and current with time.

color{blue} ✍️ We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is

color{purple}{P_C = t v = i_m cos (omegat) v_m sin (omegat)}

color{purple}{= i_mv_m cos(omegat) sin (omegat)}

color{blue}{=(i_mv_m)/2sin (2 omega t)}

.............(7.19)

color{blue} ✍️ So, as in the case of an inductor, the average power

color{purple}{P_c = (i_mv_m)/2 sin (2omegat) = (i_mv_m)/2 (sin(2omegat))=0}

color{blue} ✍️ since = 0 over a complete cycle. Figure 7.10 explains it in detail.

color{blue} ✍️ Thus, we see that in the case of an inductor, the current lags the voltage by pi/2 and in the case of a capacitor, the current leads the voltage by pi//2.
Q 3178845706

A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?
Class 12 Chapter 7 Example 3
Solution:

When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/wC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.
Q 3188845707

A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?
Class 12 Chapter 7 Example 4
Solution:

The capacitive reactance is

X_c = 1/(2pivC) = 1/(2pi(50Hz)(15.0xx10^(-6) F) = 212Omega

The rms current is

I V/(X_C) = (220V)/(212Omega) = 1.04A

The peak current is

i_m = sqrt2l = (1.411) (1.04A) = 1.47A

This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by p//2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.
Q 3108845708

A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.11. The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons
Class 12 Chapter 7 Example 5
Solution:

As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.