Alternating Current and EMF:

A current whose magnitude continuously varies with time and whose direction reverses periodically is called alternating current (AC), and the corresponding voltage is called alternating EMF. During one complete cycle, the current or EMF first rises from zero to a maximum in one direction, falls to zero, then becomes a maximum in the reverse direction, and again falls to zero.

The value of AC at any instant of time t is called instantaneous AC and is given by:

I=I₀​sinωt

Similarly, the instantaneous alternating EMF is given by:

E=E₀​sinωt

where

is the angular frequency, $T$ is the time period, I0​ is the maximum value of current called the peak value of AC, and E0​ is the maximum value of EMF called the peak value of AC EMF. The quantity ωt is called the phase of the AC current or EMF.

The variation of alternating current with time is shown in the graph. An alternating waveform that has the same shape as the sine function is called a sinusoidal waveform.

Peak Value of AC:

The maximum value of AC current in either direction is called the peak value I0​. It is also known as the current amplitude. The maximum value of AC voltage in either direction is called the peak valueE0​. It is also known as the voltage amplitude.

Mean Value or Average Value of AC

The value of AC over the positive half cycle is equal in magnitude and opposite in direction to that of the negative half cycle. Therefore, the mean value of AC over one complete cycle is zero. Hence, the average or mean value of AC is measured over any half cycle.

The average value of AC over a half cycle is defined as that value of steady (DC) current which would send the same amount of charge as sent by an AC through the same circuit in the same interval of time.

To calculate the mean value of AC, let the instantaneous value of AC passing through the circuit be represented by

I=I₀​sinωt…………………..(1)

The small amount of charge passing through the circuit in a small time $dt$ is

dq=Idt

Substituting the value of I

dq =I₀​sinωtdt

The total charge passing through the circuit in a half cycle is obtained by integrating equation (ii) from

Similarly, the mean value of ac over the negative half cycle is

Im​=−0.637I₀​

Therefore, the average ac over a complete cycle is zero. Similarly, the average value of alternating emf over a half cycle is

E_m=0.637E0​

Root mean square (RMS) value of AC/ Virtual value/ Effective value

Since the average value of ac over one complete cycle is zero, it can’t be used to specify ac current or can’t be used for power calculation. Hence more suitable quantity called RMS value of ac is defined in terms of d.c (steady current) that would do the same work (produce heat) at the same rate as that of ac do under the similar condition.

RMS value of AC is defined as the value of steady current which produce same amount of heat as produced by Ac in same resistance in the same time. Let  the instantaneous current passing through a resistance $R$ is given by,

AC through Resistor

Let a pure resistor of resistance $R$ is connected to an ac source as shown in fig. The instantaneous alternating emf of the source is given by

E=E₀​sinωt…………….(1)

Where E0​ is the peak value of emf, and $\omega$ is the angular frequency. Let $I$ be the instantaneous current in the circuit at time t.
Then,

From eq. (i) and (ii) it is obvious that the emf (voltage) $E$ and current $I$ in the resistor are in the same phase as shown in the phasor diagram.

AC through an Inductor

Let a pure inductor of inductance L is connected to an ac source as shown in fig. The instantaneous alternating emf of the source across the inductor is given by

E=E₀​sinωt…………….(1)

Where,

is the peak value of AC. From equations (i) and (ii), we find that the voltage E leads the current I by a phase angle π/2, or the current lags behind the voltage by a phase angle π/2 in an inductor. This is shown in the phasor diagram.

Inductive Reactance XL

It is the resistance offered by the inductor to the flow of alternating current in the circuit. We have,

Comparing this with,

​

It is obvious that ωL is the resistance of the inductor, called inductive reactance, and is denoted by XL Therefore,

X_L​=ωL=2πfL

Where f is the frequency of AC. In DC, $f=0$, and hence, XL=0

Its unit is ohm (Ω). The higher the frequency of AC, the greater is the inductive reactance.

AC through Capacitor

Let a capacitor of capacitance C be connected to an AC source as shown in the figure. The instantaneous alternating emf of the source across the capacitor is given by

E=E₀​sinωt…………….(1)

Where E₀ is the peak value of emf, and $\omega$ is the angular frequency. Let q be the charge stored in the capacitor at any instant t. Then the potential difference across the capacitor is

Since there is only a capacitor in the circuit, the potential difference across it is equal to the instantaneous emf of the source, i.e.

Comparing the voltage E=E0sin⁡ωt from eq. (i) with current from eq. (iii), we see that the current leads the voltage by π/2, or equivalently, the voltage lags behind the current by π/2 in a capacitor, as shown in the phasor diagram.

Capacitive Reactance X_C

Capacitive Reactance is the resistance offered by the capacitor to the flow of AC in the circuit.

We have,

where $f$ is the frequency of AC. In DC, f=0 and so X_C =∞. So a capacitor offers infinite resistance to the flow of DC, i.e., a capacitor blocks DC. Its unit is Ohm. The higher the frequency of AC, the lower is the capacitive reactance.

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AC Through Inductor and Resistor in Series (LR Series Circuit)

Let an inductor of inductance $L,$ and a resistor of resistance $R$ be connected in series with an AC source of rms emf $E$. Let $I$ be the rms value of current in the circuit. Let the potential difference across inductor $L$ be

V_L​=I X_L​ 

where X_L​=ωL is the inductive reactance.Potential difference across resistor $R$ is

V_R​=IR

Since V_R​  and I are in the same phase, V_R   is represented by OA in the direction of I. The voltage V_L​  leads the current  I by phase angle π/2, so V_L is represented by OB perpendicular to the direction of I. The resultant of VR​  and V_L ​ is represented by E=OH. The magnitude of OH is given by:

is called the impedance of the LR circuit. It is the total opposition offered by the LR circuit to the flow of AC current. If the resultant emf $E$ leads the current $I$ by an angle$\theta$, then

AC Through Capacitor and Resistor in Series (CR Series Circuit)

Let a capacitor of capacitance $C,$ and a resistor of resistance $R$ be connected in series with an AC source of rms emf $E$. Let $I$ be the rms value of current in the circuit. Potential difference across the capacitor $C$ is

V_C​=I X_C

where​,

is the capacitive reactance. Potential difference across resistor $R$ is

V_R​=IR

since V_R​  and I are in the same phase, V_R​  is represented by OA in the direction of I. The current I   leads the voltage V_C ​ by a phase angle π/2, so V_C​ is represented by OB perpendicular to the direction of  I. The resultant of V_R​  and V_C​  is represented by E=OH. The magnitude of OH is given by:

is called the impedance of the CR circuit. It is the total opposition offered by the CR circuit to the flow of AC current. If the resultant emf $E$ lags the current $I$ by an angle$\theta$, then

AC Through an Inductor, Capacitor, and Resistor in Series (LCR Series Circuit)

Description:
An inductor of inductance $L$, a capacitor of capacitance $C$, and a resistor of resistance $R$ are connected in series with an AC source of RMS emf $E$. Let $I$ be the RMS value of the current in the circuit.

An inductor of inductance L, a capacitor of capacitance C, and a resistor of resistance R are connected in series to an AC source of RMS emf E. Let I denote the RMS current in the circuit.

Potential Differences across Each Element:

Inductor L:

V_L=IX_L (Leads the current I by π/2)

Capacitor C:

V_C=IX_C (Lags the current II by π/2π/2)

Resistor R:

V_R=IR (In phase with the current II)

Phasor Representation:
Since V_R​ is in phase with I, it is represented by a phasor OA in the direction of I. Because V_L​  leads I by π/2, it is represented by OB drawn perpendicular to I in the leading (anticlockwise) direction.

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