Alternating Currents - Class 12 Physics

Introduction on AC Circuit

When a battery is connected to a circuit, the current flows steadily in one direction. It is called a direct current. Electric generators at electric power plants, however, produce alternating current due to many technical reasons. Alternating current results when a sinusoidal e.m.f or voltage is applied in a circuit. Circuits fed by the alternating source are known as A.C. circuits. Alternating current (AC) circuits carry energy due to the coordinated vibrations of neighboring electrons. While DC circuits require single electrons to (slowly) move through the circuit and carry energy thanks to the kinetic energy carried by electrons as they drift through the wire, AC manages to carry energy without any overall motion of the electrons through the circuit.

Alternating Current and E.M.F

The current or e.m.f., whose magnitude changes with time and direction reverses periodically is known as alternating current or e.m.f.

The instantaneous values of alternating current and e.m.f at any time t are given by

$$ \begin{align*} I &= I_0 \sin \omega t \dots (i) \\ E &= E_0 \sin \omega t \dots (ii) \end{align*} $$

where I₀ and E₀ are their maximum value or peak value or amplitude of current and e.m.f respectively. Their angular frequency is called driving frequency and is given by

$$\omega = \frac {2\pi }{T} = 2\pi f $$

Where T is time period and f is frequency of alternating current or e.m.f.

Advantages

1. Generation, transmission, and distribution of A.C. are more economical than that of D.C.
2. AC is easily convertible into D.C.
3. AC can be better controlled without any loss of electric power by using a choke coil.
4. The alternating high voltage can be stepped down or stepped up easily by using a transformer.
5. AC. can reach distant places without much loss of electric power by using a transformer.
6. AC. machines are easier to use.

Disadvantages

1. C. cannot be used in the electrolysis process such as electroplating, electrotyping, and electrorefining etc. where only D.C. is used.
2. C. can be more dangerous than D.C. in terms of its attractive nature and also because its maximum value is \( \sqrt 2 \) times its effective value.
3. C. in a wire is not uniformly distributed through its cross-section. The a.c. density is much greater near the surface of the wire than that inside the wire. The a.c. density is much greater near the surface of the wire than that inside the wire. The concentration of a.c. near the surface of the wire is called the skin effect.
4. Markings on the scales of A.C. meters are not equidistant for small measurement

Mean or Average Value of Alternating Current

Alternating current is positive during the first half cycle and negative during another half cycle, so the mean average value of a.c. over one cycle is zero. We can find the mean or average value of a.c. over any half cycle.

Mean or average value of alternating current is that value of steady current which sends the same amount of charge through a circuit in a certain time interval as is sent by an alternating current through the same circuit in the same time interval.

To calculate its value, let an alternating current be represented by

$$ \begin{align*} I &= I_0 \sin \omega t \\ \end{align*} $$ The charge sent by the alternating current I in time dt is given by $$ \begin{align*} \\ dq &= Idt \: \left [ \therefore I = \frac {dq}{dt} \right ] \\ dq &= I_0\sin \omega t dt \\ \end{align*} $$

The amount of charge passing through the circuit in half time period can be obtained by integrating above equation (i) from t = 0 to t = T/2.

$$ \begin{align*} q &= \int _0^{T/2} I_0 \sin \omega t \: dt = I_0 \int _0^{T/2} \sin \omega t dt \\ &= I_0 \left [ \frac {-\cos \omega t}{\omega } \right ]_0 ^{T/2} \\ &=- \frac {I_0}{\omega } [\cos \omega t ] _0^{T/2} \\ &= - \frac {I_0}{2\pi /T} \left [ \cos \frac {2\pi }{T} t \right ]_0^{T/2} \: \left [\therefore w = \frac {2\pi }{T} \right ] \\ &= - \frac {I_0 T}{2\pi } \left [ \cos \frac {2\pi }{T}\frac T2 - \cos 0 \right ] \\ &= - \frac {I_0T}{2\pi } [\cos \pi - \cos 0] \\ &=- \frac {I_0 T}{2\pi }[-1-1] \\ &= - \frac {I_0 T}{2\pi } \times -2 \\ q &= \frac {I_0 T}{\pi } \dots (ii) \\ \end{align*} $$

If I_m be the mean value of an a.c. over positive half cycle, then the charge sent by it in time T/2 is given by

$$ \begin{align*} q &= I_m \frac T2 \dots (iii) \\ \text {From equation} \: (ii) \: \text {and} \: (iii), \: \text {we get} \\ I_m \frac T2 &= \frac {I_0T}{\pi } \\\therefore I_m &= \frac {2I_0 }{\pi } \\ &= 0.637 I_0 \\ \end{align*} $$

Hence it means an average value of a.c. cover positive half cycle is 0.637 times the peak value of a.c. is 63.7% of the peak value.

Similarly, the mean or average value of a.c. over the negative half cycle is obtained by integrating equation (1) from t = T/2 to t = T. It comes out to be -0.637I₀. Hence the mean or average value of a.c. over one complete cycle is 0.637I₀ – 0.637I₀ = zero

Similarly, the mean value of alternating e.m.f is

$$ E_m = \frac {2E_0} {\pi} = 0.637 E_0 $$

Root Mean Square Value of A.C. and A.C. Through a Resistance Only

Root Mean Square (R.M.S) or Virtual Value of A.C. or Effective Value of A.C.

It is that steady current when passed through a resistance for a given time produces the same amount of heat as the alternating current does in the same resistance in the same time.

To calculate its value, let an alternating current be represented by

$$ I = I_0 \sin \omega t $$

If this alternating current flows through a resistance R for small time dt, then small amount of heat produced which is given by

$$ \begin{align*} dH &= I^2R\: dt = (I_0^2\sin^2\omega t) R\: dt \\ \text {or,} \: dH &= I_o^2 R \sin^2\omega T dt \\ \end{align*} $$

To obtain total amount og heat produced in one cycle, we have to integrate equation (i) from t = 0 to t = T.

$$ \begin{align*}&= I_o^2 R \int_0^T \sin ^2 \omega t \: dt \\ &= I_o^2 R \int _0^T \frac {(1 - \cos 2\omega t)}{2} dt \\ &= \frac {I_0^2R}{2}\int _0^T (1 - \cos 2 \omega t) dt \\ &= \frac {I_0^2R}{2} \left [ \int _0^T dt - \int _0^T \cos 2\omega t dt \right ] \\&= \frac {I_0^2R}{2} \left [ [\text {t}]_0^T - \left [\frac {\sin 2\omega t}{2\omega } \right ]_0^T\right ] \\ &= \frac {I_0^2R}{2}\left [(T-0)-\frac {1}{2\omega}\left [\sin 2\frac {2\pi}{T.}t \right ]_0^T \right ] \\ &= \frac {I_0^2R}{2} \left [ T - \frac {1}{2\omega } \left (\sin 2 \frac {2\pi }{T}T - \sin 0 \right )\right ] \\ \text {or,} \: H &= \frac {I_0^2 RT}{2} \dots (ii) \\ \end{align*} $$

If I_v be the r.m.s. value or virtual value of a.c., then the amount of heat produced in the same resistance R in same time T is written

$$ \begin{align*} H &= I_v^2 RT \dots (ii) \\ \text {From equation} \: (ii) \: \text {and} \: (iii), \text {we get} \\ I_v^2RT &= \frac {I_0^2 RT}{2} \\ \therefore \: I_v &= \frac {I_0}{\sqrt {2}} = 0.707\: I_0 \\ \end{align*} $$

Hence the r.m.s. value or virtual value or effective value of a.c. in 0.707 times the peak value of a.c. i.e 70.7% of the peak value of a.c.

The virtual value of alternating current can be proved as

$$ E_v = \frac {E_0}{\sqrt 2} = 0.707\: E_0 $$

A.C. Through a Resistance Only

An a.c. source is connected to a resistor is known as resistive circuit.

$$ \begin{align*} \text {The applied alternating e.m.f. is given by} \\ E &= E_0 \sin \omega t \\ \text {Let I be the circuit at any instant of time t. So} \\\text {The potential difference across the resistor} &= IR \\ \text {or,} \: E &= IR \\ \text {or,} \: I &= \frac ER \\ &= \frac {E_0 \sin \omega t }{R}\\ [\therefore E = E_0 \sin \omega t \: \text {in equation} \: (i)] \\\text {or,} \: I &= I_0 \sin \omega T \dots (ii)\\ \text {where} \: \frac {E_0}{R} = I_0 \: \text {is the peak value of alternating current.} \end{align*} $$

Comparing equation (i) and (ii), we find that E and I are in phase. Therefore, in an a.c. circuit containing R only, the current and e.m.f. are in the phase.

A.C. through an Inductance only

An alternative e.m.f applied to an ideal inductor of inductance L. Such circuit is known as purely inductive circuit.

$$ \begin{align*} \text {The applied alternating e.m.f is given by} \\ E &= E_0 \sin \:\omega t \dots (i) \\ \text {The induced e.m.f across the inductor} = -L \frac {dI}{dt} \\ \text {which opposes the growth of current in the circuit.} \\ \text {As there is no potential drop across the circuit.} \\ \text{ so, we can write,} \\ E + \left (-L \frac {dI}{dt}\right ) = 0 \\ \text {or,} \: L \frac {dI}{dt} &= E \\ \text {or,} \: \frac {dI}{dt} &= \frac EL = \frac {E_0}{L} \sin \omega t \\ \text {or,}\: dI &= \frac {E_0}{L} \sin \: \omega t \: dt \\ \text {integrating both sides, we get} \\ \int dI &= \int \frac {E_0}{L} \sin \omega t \: dt \\ \text {or,} \: I &= \frac {E_0}{L} \int \sin\omega t \: dt \\ &= \frac {E_0}{L} \left (-\frac {\cos \omega t}{\omega } \right ) = \frac {E_0}{L\omega } (-\cos \: \omega t) \\ \text {or,} \: I &= \frac {E_0}{L\omega } \sin (\omega t - \pi /2) \\ [\because \sin(\omega t - \pi /2) = - \cos \omega t] \\ \therefore I &= I_0 \sin (\omega t - \pi /2) \dots (ii) \\ \text {where} \frac {E_0}{L\omega } = I_0 \: \text {is peak value of a.c.} \\ \end{align*} $$

Comparing equation (i) and (ii), we find that in an a.c. circuit containing L only, current I lag behind the voltage E by a phase angle of 90^o.

Inductive Reactance (X_L)

Comparing \( I_0 = \frac {E_0}{L\omega } \) with \(I_0 = \frac {E_0}{R} \), we conclude that \((L\omega )\) has the dimension of resistance. The term \((L\omega )\) is known as inductive reactance represented by X_L.

The inductive reactance is the effective opposition offered by the inductor to the flow of current in the circuit.

Thus, \( X_L = \omega L = 2\pi fl \)

where f is frequency of a.c. supply.

In d.c. circuits, f = 0

$$ \therefore X_L = 0 $$

i.e A pure inductance offers zero resistance to d.c. further, \(X_L \propto f\), i.e. higher the frequency of a.c., more is the inductive reactance.

$$ \begin{align*} \text {i.e.} \: X_L &= \omega L \rightarrow \frac {1}{\text {sec}} \: (\text {henry}) \\ &= \frac {1}{\text {sec}} \frac {\text {volt}}{\text {amp}/\text {sec}} &= \text {ohm} \\ \end{align*} $$

A.C. through a Capacitor Only

The applied alternating e.m.f. across the capacitor is given by

$$ \begin{align*} E &= E_0 \: \sin \omega t \dots (i) \\ \text {Let q be the charge on the capacitor at any instant.} \\ \text {So, potential difference across the capacitor,} \: V = \frac qc \\ \text {or,} \: E &= \frac qc [\therefore V = E] \\ \text {or,} q &= E.C. = E_0 \: \sin \: \omega t \: \\ \text {or,} I &= \frac {dq}{dt} = \frac {d}{dt} (E_0 \: \sin \: \omega t c ) \\ \text {or,} I &= E_0 \: c\: (\sin \omega t \: c) \\ \text {or,} \: I &= E_0 \: c (\cos \omega t) \omega \\ \text {or,} \: I &= \frac {E_0}{1/c\omega }\cos \omega t \\ \text {or,} \: I &= \frac {E_0}{1/c\omega} \sin\: (\omega t + \pi/2) \\ \therefore I &= I_0 \sin (\omega t + \pi/2) \dots (ii) \\ \text {where} I = \frac {E_0}{1/c\omega} = I_0 \: \text {is peak value of a.c.} \\ \end{align*} $$

Capacitive Reactance (X_c)

Comparing \( I_0 \frac {E_0}{1/c\omega } \) with \(I_o = \frac {E_o}{R} \), we conclude that \(\left (\frac {1}{c\omega } \right )\) has the dimension of resistance. The term \(\left (\frac {1}{c\omega } \right ) \) is known as capacitive reactance \(X_c\). The capacitive reactance is the effective opposition offered by a capacitor to the flow of current in the circuit. Its unit is ohm in SI-system.

$$ \begin{align*} \text {Thus,} \: \: \: X_C &= \frac {1}{\omega c} = \frac {1}{2\pi fc} \\ \text {where f is frequency of a.c. supply} \\ \text {In a d.c. circuit, }\: f= 0, \\ \therefore X_c &= \infty \\ \text {i.e. a condenser will block d.c.} \\ \text {The unit of} \: X_c\: \text {can be deduced as:} \\ X_c &= \frac {1}{\omega c} = \text {sec} \frac {1}{\text {farad}} \\ &= \text {sec} \frac {1}{\text {coloumb}/ \text {volt}} = \frac {\text {volt sec}}{\text {Amp sec}} = \text {ohm} \\ \text {Hence} \: X_c \: \text {is measured in ohm.} \\ \end{align*} $$

A.C. through an Inductance and Resistance

Suppose a pure resistance R and a pure inductance L is connected in series to a source of alternating e.m.f. shown in a figure. Let E be the r.m.s value of applied alternating e.m.f. and I be the r.m.s value of current flowing in the circuit.

The potential difference across the inductor, \(V_L = IX_L \) (leads current I by an angle of p/2).

The potential difference across R, \(V_R = I.R. \) (in phase with the current).

Since V_R and I are in phase so V_R is represented by OA in the direction of I. the current lags behind the potential difference V_L by angle of \(\pi /2\) so V_L is represented by OB perpendicular to the direction of I. So the resultant of V_R and V_L is given by OH. The magnitude of OH is given by

$$ \begin{align*} OH &= \sqrt {OA^2 + OB^2} = \sqrt {V_R^2 + V_L^2} \\ \text {or,} \: E &= \sqrt {I^2R^2 + I^2X_L^2} \\ \text {or,} \: E &= I\sqrt {R^2 + X_L^2} \\ \text {or,} \: \frac EI &= \sqrt {R^2 + X_L^2} \\\end{align*} $$

But \(\frac EI = z\), is effective opposition of L-R circuit to a.c. called impedance of LR circuit. The impedance of L-R circuit is given by

$$ \begin{align*} Z &= \sqrt {R^2 + X_L^2} \\ \text {Again,} \\ I &= \frac {E}{Z} = \frac {E}{\sqrt {R^2 + X_L^2}} \\ \therefore I &= \frac {E}{\sqrt {R^2 + (L\omega )^2}} \:\:\: [ \therefore X_L = L\omega ] \\ \text {Let} \:\theta \: \text {be the angle between E and I, so from figure, } \\ \text {we have}, \\ \tan \theta &= \frac {V_L}{V_R} = \frac {IX_L}{IR} \\ \text {or,} \: \tan \: \theta &= \frac {X_L}{R} = \frac {L\omega }{R} \\ \therefore \tan \: \theta &= \frac {L\omega }{R} \\ \end{align*} $$

If the values of X_L and R are known as θ can be calculated. Current lags behind applied voltage or e.m.f.

A.C. through Capacitance and a Resistance

Let a pure resistance R and an ideal capacitor of capacitance C be connected in series to the source of alternating e.m.f shown in a figure. Let E be the r.m.s value of applied alternating e.m.f and I be the r.m.s value of current flowing in the circuit. The potential difference across R, \(V_R = IR\) (in phase with the current).

The potential difference across C is \( V_C = I X_c \) (lags behind the current by I by angle \(\pi /2\).

Since V_R and V_C is given by OH. The magnitude of OH is given by

$$ \begin{align*} OH &= \sqrt {OA^2 + OB^2} \\ \text {or,} \: E &= \sqrt {V_R^2 + V_C^2} = \sqrt {(IR)^2 + (IX_c)^2} \\ \text {or,} \: E &= I \sqrt {R^2 + X_c^2} \\ \text {or,} \: I &= \frac {E}{\sqrt {R^2 + X_c^2}} \\ \text {or,} \: \frac EI &= \sqrt {R^2 + X_c^2} \\ \end{align*} $$

But E/I = z, is the effective opposition of C-R circuit to a.c. called impedance of C-R circuit. The impedance of C-R circuit is given by

$$ \begin{align*} Z &= \sqrt {R^2 + X_c^2 } \\ I &= \frac EZ = \frac {E}{\sqrt {R^2 + X_c^2}} \\ \therefore I &= \frac {E}{R^2 + ( \frac {1}{\omega c})^2} \\ \text {Let} \: \theta \: \text {be the angle between E and I} \\ \text { So from figure, we have,} \\ \tan \: \theta &= \frac {-V_c}{V_R} = \frac {-IX_c}{IR} \\ \tan \theta &= \frac {X_c}{R} \\ \end{align*} $$

Since current is taken as the reference phasor, negative phase angle implies that voltage lags behind the current. It is same as current leads the voltage or emf.

A.C. through an Inductance, Capacitance and a Resistance

Let a pure resistance R, a pure inductance L and an ideal capacitor of capacitance C be connected in a series to a source of alternating e.m.f. shown in a figure. As R, L and C are in series, current at any instant through the three elements has the same amplitude and phase. However, a voltage across each element bears different phase relationship with the current. Let E be the r.m.s. the value of the applied alternating e.m.f. to LCR circuit and I be the r.m.s value of current flowing through all the circuit elements.

Points to be noted are:

1. The potential difference across inductor, \(V_L = IX_L \)
   (leads current I by an angle of \(\pi /2\))
2. The potential difference across C, \(V_c = IX_c\)
   (leads behind the current I by an angle of \(\pi /2\))
3. The potential difference across R, \(V_R = IR \)
   (in phase with the current)
4. Since V_R and I are in phase so V_R is represented by OA in the direction of I as shown in the figure.
5. The current lags behind the potential difference V_L by an angle of \(\pi /2\), so V_L is represented by OB perpendicular to the direction of I.
6. The current leads the potential difference V_C by an angle of \(\pi /2\) so V_c is represented by OF perpendicular to the direction of I.
7. Since V_L and V_C are in opposite phase, so their resultant \((V_L – V_C)\) is represented by OD (Here V_L > V_C).

The resultant of V_R and \((V_L – V_C)\) is given by OH. The magnitude of OH is given by

$$ \begin{align*} OH &= \sqrt {(OA)^2 + (OD)^2} = \sqrt {V_R^2 + (V_L – V_C)^2} \\ \text {or,} \: E &= \sqrt {I^2R^2 + (IX_L – IX_C)^2} \\ &= I\sqrt {R^2 + (IX_L – IX_C)^2} \\ \text {or,} \: \frac EI &= \sqrt {R^2 +(X_L – X_C)^2} \\\end{align*} $$  But\( \: E/I = Z \) is the effective opposition of LCR circuit to A.C. called impedance of the circuit. $$ \begin{align*} \text {So, we have} \\ \ Z &= \frac EI \\ &= \sqrt {R^2 +(X_L – X_C)^2} \dots (i) \\ \text {Let} \: \theta \: \text { be the phase angle between E and I,} \\ \text {so from figure, we have} \\ \tan \theta &= \frac {AH}{OA} \\ &= \frac {V_L – V_C}{V_R} \\ \frac {IX_L – I X_C}{IR} &= \frac {X_L – X_C}{R} \\ \therefore \tan \theta &= \frac {\left (L\omega - \frac {1}{C\omega } \right )}{R} \\ \end{align*} $$

The equation (i) is the general relation for impedance.

1. When \( X_L = X_C \), then \(\tan \theta = 0\) or \( \theta = O^o\). Hence voltage and current are in same phase.
2. When \( X_L > X_C \), then \(\tan \theta \) or \(\theta \) is positive. Hence voltage leads the current by phase angle \(\theta \). The a.c. circuit is inductance dominated circuit.
3. When \( X_L < X_C \), then \(\tan \theta \) or \(\theta \) is negative. Hence current leads the voltage by phase angle \(\theta \). The a.c. circuit is capacitor dominated circuit.

Impedance and Admittance

The total effective opposition offered by LCR circuit to alternating current is known as impedance and is denoted by z. The reciprocal of the impedance of a circuit is known as an admittance of the circuit.

$$ \text {i.e.} \: \: \: \text {Admitance} \:(A) = \frac 1Z $$

Unit of impedance (Z) of the circuit is ohm.

Unit of admittance of the circuit is ohm^-1 i.e. mho or siemen.

Electrical Resonance in Series LCR Circuit

Electrical resonance is said to take place in a series LCR circuit when the circuit allows maximum current for a given frequency of the source of alternating supply for which capacitive reactance becomes equal to the inductive reactance.

The current (I) in a series LCR circuit is given by

$$ \begin{align*} I &= \frac EZ = \frac {E}{\sqrt {R^2 + \left (l\omega - \frac {1}{C\omega }\right )^2}} \end{align*} $$

At high frequency, \(X_L = L\omega \) is very large and \(X_C = \frac {1}{C\omega } \) is very small.

At low frequency, \( X_L = L\omega \) is very small and \(X_C = \frac {1}{C\omega} \) is very large.

If \(X_C = X_L \) for a particular frequency f₀, then the impedance of LCR circuit is given by

$$ Z = \sqrt {R^2 + 0} = R $$

Impedance of LCR circuit is minimum and hence current is maximum. This frequency F₀ is called electrical resonance.

Determination of Resonant Frequency

For electrical resonance, we have

$$ \begin{align*} X_L &= X_C \\ \text{or,} \: L\omega &= \frac {1}{C\omega } \\ \text {or,} \: \omega ^2 &=\frac {1}{LC} \\ \text {or,} \: \omega &= \frac {1}{\sqrt {LC}} \\ \text {or,} \: 2\pi f_o &= \frac {1}{\sqrt {LC}} \\ \therefore f_o &=\frac {1}{2\pi \sqrt {LC}} \\ \end{align*} $$

Application of LCR Circuit

At resonance of LCR circuits admit maximum current at particular frequencies. LCR circuit is used in transmitters and receivers of radio, television and telephone carrier equipment etc.

Quality Factor of Resonance Circuit

The quality factor or Q-factor of a series resonant circuit is defined as the ratio of a voltage developed across the inductance or Capacitance at resonance to the impressed voltage, which is the voltage applied across R.

Q-factor by inductor

$$ \begin{align*} \text {i.e.} Q &= \frac {\text {Voltage across L}}{\text {applied voltage}} \\ \text {or,} \: Q &= \frac {IX_L}{IR} \\ \text {or,} \: Q &= \frac {\omega L}{R} \\ \text {or,} \: Q &= \frac {1}{\sqrt {LC}}.\frac LR \\ \text {or,} \: Q &= \sqrt {\frac LC}. \frac 1R \\ \text {or,} \: Q &= \frac 1R \sqrt {\frac LC} \\ \end{align*} $$

Q-factor by capacitor

$$ \begin{align*} \text {i.e.} Q &= \frac {\text {Voltage across C}}{\text {applied voltage}} \\ \text {or,} \: Q &= \frac {IX_C}{IR} \\ \text {or,} \: Q &= \frac {1 }{\omega C R} \\ \text {or,} \: Q &= \frac {1}{\frac {1}{\sqrt {LC}}C.R} \\ \text {or,} \: Q &=\frac{ \sqrt { LC}}{CR} \\ \text {or,} \: Q &= \frac 1R \sqrt {\frac LC} \\ \end{align*} $$

Power consumed in a Series LCR Circuit

Let the alternating e.m.f. applied to LCR circuit be

\begin{align*} E &= E_0\sin\: \omega t \dots (i) \\ \end{align*}

If the alternating current developed lags behind the applied e.m.f. by a phase angle $$ \begin{align*}\: \theta , \\ \text {then} \\ I &= I_0 \sin (\omega t - \theta ) \dots (ii) \\ \text {Power at instant t is given by} \\ \frac {d\omega }{dt} &= EI \\ &= E_0\sin\: \omega t \times I_0 \sin (\omega t - \theta ) \\ &= E_0I_0 \sin \omega t \:(\sin \omega t \cos \theta - \cos \omega t \sin \theta ) \\ &= E_0I_0 \sin^2\omega t \cos \theta – E_0I_0\sin \omega t \: \cos \: \omega t\sin \: \theta \\ &= E_0I_0 \sin ^2\omega t \cos \theta - \frac {E_0I_0}{2} \sin \: 2\omega t\:\sin \: \theta \\ \end{align*} $$

If this instantaneous power is assumed to remain constant for a small constant for a small time dt, then small amount of work done in this time is

$$ \begin{align*} dW = \left ( E_0I_0 \sin ^2 \omega t \cos \theta - \frac {E_0I_0}{2} \sin \: 2\omega t \sin \: \theta \right )dt \\ \end{align*} $$

Total work done over a complete cycle is obtained by integrating above equation from 0 to T.

$$ \begin{align*} \text {i.e.} \: W &= \int _0^T E_0I_0 \sin^2 \omega t \: \cos \theta \: dt - \int _0^T \frac {E_0I_0}{2} \sin \: 2\omega T \: dt \dots (iii) \\ \text {or,} \: W &= E_0I_0 \: \cos \:\theta \int _0^T \sin ^2 \omega t \: dt - \frac {E_0I_0}{2} \sin \: \theta \:\int _0^T \sin \: 2\: \omega t \: dt \\ \text {Now,} \\\end{align*} $$

$$ \begin{align*} \int _0^T \sin^2\omega t \: dt &= \int _0^T \left (\frac {1-\cos \: 2\omega t}{2} \right )dt \\ &= \frac 12 \left [ \int _0^T dt - \int _0^T \cos \:2\omega t \: dt \right ] \\ &= \frac 12 [T -0] \\ &= \frac T2 \dots (iv) \\ \text {and} \: \int _0^T \sin \: 2\omega t \: dt &= 0 \dots (v) \\ \text {Using equation} \: (iv) \: \text {and} \: (V) \: \text {in equation} \: (iii), \\ \text {we get} \\ \end{align*} $$

$$ \begin{align*} W &= E_0I_0 \: \cos \: \theta \frac T2 – 0 \\ W &= \frac {E_0I_0 \cos \: \theta}{2} \\ \therefore \end{align*} average power in the inductive circuit over a complete cycle\begin{align*} \\ P &= \frac WT \\&= \frac {E_0I_0 \cos \: \theta }{T} \frac T2 \\ &= \frac {E_0}{\sqrt 2} \frac {I_0}{\sqrt 2} \cos \: \theta \\ P &= E_vI_v \: \cos \theta \dots (iv) \\ \end{align*} $$

Hence, average power over a complete cycle in an inductive circuit is the product of virtual e.m.f., virtual current and cosine of the phase angle between the voltage and the current. The quantity \(\cos \theta \) is called power factor.

Here, P is called true power and E_vI_v is called apparent power or virtual power.

Special cases:

1. A.C. circuit containing resistor only:
   In such a circuit, phase angle, \(\theta = 0^o\), so true power dissipated \((P) = E_vI_v \cos \:0^o\: =E_vI_v\)
   $$ \begin{align*}\text {or,} \: \text {True power} &= \text {apparent power} \\ \text {Power loss} &= \text {Product of virtual value of voltage and current} \\ \end{align*} $$
2. A.C. circuit having pure inductor or capacitor only:
   In such a circuit, the phase angle, between voltage and current is \(\frac {\pi}{2} \).

$$ \begin{align*} \theta &= \frac {\pi }{2} \\ \therefore \text {Power dissipated, P} = E_vI_v\: \cos \left (\frac {\pi }{2} \right ) = 0 \\ \end{align*} $$
Thus, no power loss takes place in a circuit having pure inductor or capacitor only.

Power Factor of an A.C. Circuit

The ratio of true power and apparent power in an a.c. circuit is called power factor of the circuit.

$$ \begin{align*}\text {i.e. Power factor,}\: \cos \theta = \frac {P}{E_vI_v} = \frac {P}{P_v} \\ \end{align*} $$

Power factor \(cos \: \theta \) is always positive and not more than 1.

1. For circuit having pure reistance, \(\cos \: \theta = 1\)
2. For circuit having pur inductor or pure capacitor, \(\cos \: \theta = 0\: \:\: [\because \theta = \frac {\pi }{2}]\)
3. For RC circuit, \(\cos \: \theta = \frac {R}{\sqrt {R^2} + \frac {1}{\omega ^2C^2}}\)
4. For LR circuit, \(\cos \: \theta = \frac {R}{\sqrt {R^2 + L^2\omega ^2}}\)
5. For LCR circuit, \(\cos \: \theta = \frac {R}{\sqrt {R^2 + \left ( L\omega - \frac {1}{C\omega }\right )^2}} \)

Wattless and Wattful Current

In pure inductor or an ideal capacitor, \(\theta = 90^o\), so average power consumed in a pure inductor or capacitor, \(P = E_vI_v \cos 90^o = 0^o\).

The current through pure inductor or capacitor consumes no power for its maintenance in the circuit is called idle wattles current.

At resonance, \( X_L = X_C \) and \(\theta = 0^o \).

$$\therefore \cos \: \theta = \cos \: 0^o = 1$$

In this case, maximum power is dissipated. The current through resistance R which consumes power for its maintenance in the circuit is called wattful current.

Choke Coil

In a d.c. circuit, current can be reduced by using suitable resistance. While using resistance, electrical power is lost in the form of heat across the resistance. But in a case of a.c. circuit, choke coil can be used to reduce current without losing power by it. The reason is alternating e.m.f leads the current by phase angle \(\pi/2\) and average power consumed will be

$$P = E_vL_V \: \cos \: \pi/2 = 0$$

So an inductance used to control current in an a.c. a circuit without much loss of energy is called choke coil.

 

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