# Alternating Current (AC) Class 12 Physics | Note

### Alternating current

A type of current or voltage whose magnitude changes with time and direction reverse periodically is known as alternating current or alternating voltage.

The instantaneous value of alternating current is

I  =  Io sin $\omega t$.

Where Io is the maximum (peak) value of current.

The instantaneous value of alternating voltage is

E  =  Eo sin$\omega t$.

Where Eo is maximum (peak) value of voltage (emf)

### Measurement of alternating current

1. The average value of alternating current measured in complete cycle is zero.

2. The average value of alternating current measured in half cycle is known as average value. The average value of alternating current is

IAvg.  =  0.637 Io

This value is not in good agreement with practical value.

3. Root mean square value of current (Ir.m.s)

Root Mean Square value of current is defined as the steady current (D.C.) which develops heat in a given resistance (R) for a given time (t) as that done by alternating current for the same resistance in same time. It is denoted by Irms.

Let us consider a resistor of resistance R is connected with an AC source. Let I = Io sin$\omega t$ be the instantaneous value of current flowing through the resistance. The small amount of work done by the alternating current in small time dt is

dw  =  I2 Rdt

or, dw  =  Io2 sin2 $\omega t$ Rdt

or, dw  =  Io2 R sin2$\omega t$ dt ……………… (i)

The total amount of work done in a complete cycle can be obtained by integrating equation (i) from limit O to T.

W  =$\int\limits_{0}^{T}{{}}$Io2 R sin2$\omega t$ dt

or, W  =  Io2 R $\int\limits_{0}^{T}{{}}$  sin2$\omega t$ dt

Or, I  =$\frac{{{E}_{o}}}{L}$$\left( -\frac{\cos \omega t}{\omega } \right) Or, I = – \frac{{{E}_{o}}}{\omega L} cos\omega t Or, I = – \frac{{{E}_{o}}}{{{X}_{L}}} cos\omega t (Where XL = \omega L is inductive reactance i.e. total resistance offered by inductor.) Or, I = – Io cos\omega t Or, I = – Io sin\left( \frac{\pi }{2}-\omega t \right) Or, I = Io sin\left( \omega t-\frac{\pi }{2} \right) Since we have started from E = Eo sin\omega t and arrived at I = Io sin\left( \omega t-\frac{\pi }{2} \right), the wave and phase diagram are; ### AC circuit containing capacitor only. Fig: AC circuit containing capacitor only. Let us consider a capacitor of capacitance C is connected with an AC source of emf E = Eo sin\omega t. Let Vc be the potential across the capacitor. From Kirchhoff’s second law (KVL), E = Vc Or, Vc = Eo sin\omega t Let us consider a small amount of charge (dq) flows in small time (dt). Now, the current is I = \frac{dq}{dt} Or, I = \frac{d({{V}_{c}}C)}{dt} since, q = VC Or, I = \frac{d}{dt} ( C Eo sin\omega t ) Or, I = Eo C \frac{d}{dt}( sin\omega t ) Or, I = Eo C (\omega \cos \omega t) Or, I = Eo\omega Ccos\omega t Or, I = \frac{{{E}_{o}}}{1/\omega C} cos\omega t Or, I = \frac{E{}_{o}}{{{X}_{L}}} cos\omega t (Where Xc = \frac{1}{\omega c} is capacitive reactance i.e. resistance offered by capacitor in a.c. circuit) Or, I = Io cos\omega t Or, I = Io sin\left( \omega t+\frac{\pi }{2} \right) Since, we have started from E = Eo sin\omega t and arrived at I = Io sin\left( \omega t+\frac{\pi }{2} \right), the wave and phase diagram are; ### AC circuit containing resistor and inductor in series. Fig: AC circuit containing resistor and inductor in series. Let us consider a resistor of resistance R and an inductor of inductance L are connected in series with an AC source of emf E = Eo sin\omega t. Let VP and VL be the potential across resistor and inductor respectively. In AC circuit containing resistor only the current I and voltage VR will in the same phase. The voltage VR is reactant by OA in the figure. In an AC circuit containing an inductor only, the voltage leads the current by phase angle \frac{\pi }{2} . The voltage VL is represented by OC in the figure. From \Delta OAB, (OB)2 = (OA)2 + (AB)2 Or, (OB)2 = (OA)2 + (OC)2 Or, VRL2 = VR2 + VL2 Or, VRL2 = I2R2 + I2XL2 ( XL = Inductive reactance) Or, VRL2 = I2 (R2 + XL2) Or, \frac{{{V}_{RL}}^{2}}{{{I}^{2}}} = R2 + XL2 Or, Z2 = R2 + XL2 (where Z = \frac{{{V}_{RL}}}{I} is known as impedance i.e. total resistance offered by a.c. circuit.) \therefore Z = \sqrt{{{R}^{2}}+{{X}_{L}}^{2}} Calculation of phase angle (\phi ): In figure, ∠AOB =\phi is the phase angle. From \Delta OAB Tan\phi = \frac{AB}{OA} = \frac{OC}{OA} Or, Tan\phi = \frac{{{V}_{L}}}{{{V}_{R}}} Or, Tan\phi = \frac{I{{X}_{L}}}{IR} Or, Tan\phi = \frac{{{X}_{L}}}{R} \therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{L}}}{R} \right) The above result shows that voltage leads the current by phase angle\phi , i.e. current lags behind the voltage by phase angle\phi . ### AC circuit containing resistor and capacitor in series. Let us consider a capacitor of capacitance C and a resistor of resistance R are connected in series with an AC source of emf E = Eo sin\omega t. Let VR = IR and VC = IXL be the potential across resistor and capacitor, respectively. When an AC circuit is connected with a resistor only, the current I and voltage VR will be in the same phase. The voltage VR represented by OA in the figure. When an AC circuit is connected with a capacitor only, the voltage lags behind the current by phase angle \frac{\pi }{2} . The voltage VC is represented by OC. Let us complete a parallelogram OABC. Now, the diagonal OB represents effective voltage E (i.e. VRC). From \Delta OAB, (OB)2 = (OA)2 + (AB)2 or, (OB)2 = (OA)2 + (OC)2 or, E2 = VR2 + VC2 or, E2 = (IR)2 + (IXC)2 or, \frac{{{E}^{2}}}{{{I}^{2}}} = R2 + XC2 or, Z2 = R2 +XC2 (where Z = \frac{E}{I} is impedance total resistance offered by a.c. circuit.) \therefore Z =\sqrt{{{R}^{2}}+{{X}_{C}}^{2}} Calculation of phase angle: From \Delta OAB, Tan\phi = \frac{AB}{OA} = \frac{OC}{OA} or, Tan\phi = \frac{{{V}_{C}}}{{{V}_{R}}} or, Tan\phi = \frac{I{{X}_{C}}}{IR} or, Tan\phi = \frac{{{X}_{C}}}{R} \therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{C}}}{R} \right) The above result shows that in an AC circuit containing capacitor and resistor in series, voltage lags behind the current by phase angle\phi . ### AC circuit containing inductor, capacitor and resistor in series. (LCR in series). Let us consider 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 emf, E = Eo sin\omega t. Let VL, VC and VR be the potential across inductor, capacitor and resistor, respectively. In AC circuit containing resistor only, the current and voltage will be in the same phase. The voltage VR is represented by OA in the figure. In AC circuit containing inductor only, the voltage leads the current by phase angle \frac{\pi }{2} , the voltage VL is represented by OY’ in the figure and in AC circuit containing capacitor only, the voltage lags behind the current by phase angle \frac{\pi }{2}, the voltage VC is represented by OY in the figure, The voltage VL and VC are in opposite direction. If (VL > VC) then the resultant voltage of VL and VC i.e., (VL – VC) is represented by OC. Let us complete a parallelogram OABC. Now the diagonal OB represents the effective voltage E of the LCR series circuit. From \Delta OBA, (OB)2 = (OA)2 + (AB)2 or, (OB)2 = (OA)2 + (OC)2 or, E2 = VR2 + (VL – VC)2 or, E2 = I2R2 + (I XL – I XC)2 or, \frac{{{E}^{2}}}{{{I}^{2}}} = R2 + (XL – XC)2 or, Z2 = R2 + (XL – XC)2 (⸪ Z = \frac{E}{I} is impedance of AC circuit i.e. total resistance offered by AC circuit) \therefore Z = \sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}} Calculation of phase angle: From \Delta OAB, Tan\phi = \frac{AB}{OA} = \frac{OC}{OA} or, Tan\phi = \frac{{{V}_{L}}-{{V}_{C}}}{{{V}_{R}}} or, Tan\phi = \frac{I{{X}_{L}}-I{{X}_{C}}}{IR} or, Tan\phi = \frac{{{X}_{L}}-{{X}_{C}}}{R} \therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right) Special cases: I. If XL > XC In this case, the value of phase angle\phi is positive i.e. voltage leads the current by phase angle\phi . II. If XL = XC In this case, the value of phase angle\phi is O. i.e. voltage and current will be in the same phase. III. If XC > XL In this case, the phase angle\phi is negative i.e. voltage lags behind the current by phase angle\phi . ### Power consumed by AC circuit (LCR in series): Fig: AC circuit containing inductor, capacitor and resistor in series Let an inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series with an AC source. Let I and E be the rms value of current and emf, respectively. Let us consider the case in which current lags behind the voltage by phase angle\phi . The instantaneous power developed is Pins = I.E. or, Pins = Io sin (\omega t –\phi ) . Eo sin\omega t or, Pins = Io Eo (sin\omega t. cos\phi – cos\omega t sin\phi ) sin\omega t or, Pins = Io Eo (sin2\omega t. cos\phi – sin\omega t cos\omega t sin\phi ) Let dw be the small amount of work done in small time dt i.e. dw = Pins . dt ………………… (i) Now, the total amount of work done in a complete cycle can be obtained by integrating equation (i) from 0 to T. or, \int\limits_{0}^{T}{{}} dw = \int\limits_{0}^{T}{{}} Pins . dt or, W = I. E. \int\limits_{0}^{T}{{}} (sin2\omega t cos\phi – sin\omega t cos\omega t . sin\phi ) dt or, W = I. E.\left[ \cos \phi \int_{0}^{T}{\left( \frac{1-\cos 2\omega t}{2} \right)dt-\frac{\sin \phi }{2}\int_{0}^{T}{\sin 2\omega t.dt}} \right] or, W =\frac{{{I}_{o}}E{}_{o}}{2}$$\left[ \cos \phi \left( \int_{0}^{T}{dt-}\int_{0}^{T}{\cos 2\omega tdt} \right)-\sin \phi .0 \right]$        $\left( \because \text{ }\int\limits_{0}^{T}{\sin 2\omega t\text{ dt =0}} \right)$

or, W  =$\frac{{{I}_{o}}E{}_{o}}{2}$  [cos$\phi$ {T – O}]   $\left( \because \text{ }\int\limits_{0}^{T}{\cos 2\omega t\text{ dt =0}} \right)$

or, W  =$\frac{{{I}_{o}}E{}_{o}}{2}$ . cos$\phi$ . T

or, W  =  $\frac{{{I}_{o}}}{\sqrt{2}}$  .  $\frac{{{E}_{o}}}{\sqrt{2}}$  cos$\phi$ T

or, W  =  Irms . Erms  cos$\phi$ T

Now, the average power developed in complete cycle is

Pavg  =  $\frac{W}{T}$

$\therefore$    Pavg  =  Irms Erms . cos$\phi$

Where, cos$\phi$  =  $\frac{R}{Z}$ is known as power factor.

Special cases

I. Resistor only

We have, Pavg.  =  Irms Erms cos$\phi$

or, Pavg.  =  Irms Erms cos O  ($\phi$ = 0°)

or, Pavg.  =  Irms Erms

II. Inductor only

We have, Pavg.  =  Irms Erms cos$\phi$

or, Pavg.  =  Irms Erms cos $\left( \frac{-\pi }{2} \right)$  $\left( \because \phi =\frac{-\pi }{2} \right)$

$\therefore$ Pavg.  =  0

III. Capacitor only

We have, Pavg.  =  Irms . Erms cos$\phi$

or, Pavg. = Irms . Erms cos$\left( \frac{\pi }{2} \right)$     $\left( \because \phi =\frac{\pi }{2} \right)$

or, Pavg.  =  0

IV. L and R in series

We have,  Pavg.  =  Irms Erms cos$\phi$

cos$\phi$ =  $\frac{R}{Z}$

Z  =  $\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$

V. R and C in series

We have, Pavg. = Irms  Erms cos$\phi$

cos$\phi$  =  $\frac{R}{Z}$

Z  =$\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}$

### Electrical resonance:

In an AC circuit containing L, C and R in series if XL = XC, the value of impedance becomes minimum and current becomes maximum. This condition is known as electrical resonance. In electrical resonance, current and voltage will be in same the phase.

For electrical resonance,

XL = XC

or, $\omega L$  =  $\frac{1}{\omega C}$

or, 2$\pi$fo L  =  $\frac{1}{2\pi {{f}_{o}}C}$

or, fo2  =  $\frac{1}{4{{\pi }^{2}}LC}$

or, fo  =  $\frac{1}{2\pi }$  $\sqrt{\frac{1}{LC}}$

Where fo is known as resonance frequency.

### Quality factor:

Quality factor is defined as the ratio of potential difference across inductor or capacitor to the potential difference across resistor in resonance condition.

Q. factor =  $\frac{P.d.\text{ }across\text{ }inductor}{P.d.\text{ }across\text{ }resistor}$

=  $\frac{{{V}_{L}}}{V{}_{R}}$

=  $\frac{I{{X}_{L}}}{IR}$

=  $\frac{\omega L}{R}$

=  $\frac{2\pi {{f}_{o}}L}{R}$

= $2\pi .\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\frac{L}{R}$

Q. factor =  $\frac{1}{R}$  $\sqrt{\frac{L}{C}}$

OR

Q. factor = $\frac{P.d.\text{ }acrossc\text{ }apacitor}{P.d.\text{ }across\text{ }resistor}$

=  $\frac{{{V}_{C}}}{{{V}_{R}}}$

=  $\frac{I{{X}_{C}}}{IR}$

=  $\frac{1/\omega C}{R}$

=  $\frac{1}{2\pi {{f}_{o}}CR}$

=  $\frac{1}{2\pi .\frac{1}{2\pi }\sqrt{\frac{1}{LC}}.CR}$

Q. factor =  $\frac{1}{R}$ $\sqrt{\frac{L}{C}}$

### Choke coil:

A coil having a large number of turns and negligible internal resistance is known as a choke coil. It is a pure inductor.

Average power consumed in AC circuit containing inductor only is

Pavg.  =  Irms  Erms cos$\phi$

$\because$ For inductor only,$\phi$ =  $\frac{\pi }{2}$

PAvg.  =  Irms  Erms cos $\frac{\pi }{2}$

$\therefore$ PAvg.  = O

### Wattless current:

In an AC circuit containing an inductor only or capacitor only, the value of phase angle is $\frac{\pi }{2}$ and average power consumed is O. Current in this case is known as wattles current.

### Wattful current:

In an AC circuit containing resistor only, the average power consumed is Irms Erms. The current in this case is known as watt full current.

Admittance is defined as the reciprocal of impedance. It is denoted by Y.

Mathematically,

Admittance, Y  =  $\frac{1}{Impedance\text{ }(Z)}$

Y  =  $\frac{1}{\sqrt{R2+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}$

Its unit is  $\Omega$–1.

Alternating Current Formulae for Numerical Problems:

1. The instantaneous value of alternating current is

I  =  Io sin$\omega t$.

Where Io is maximum (peak) value of current.

And the instantaneous value of alternating voltage is

E  =  Eo sin$\omega t$.

Where Eo is maximum (peak) value of voltage.

2. For certain phase angle ($\phi$)

E  =  Eo sin$\omega t$ and I = Io sin ($\omega t\pm \phi$) where φ is phase angle between emf and current.

3. Irms  =  $\frac{{{I}_{o}}}{\sqrt{2}}$ & Erms  =  $\frac{{{E}_{o}}}{\sqrt{2}}$

4. Phase angle ($\phi$) between current and voltage in AC circuit containing

5. Inductive reactance (i.e. resistance offered by inductor),

XL =$\omega L$ = 2$\pi$f L

Capacitive reactance (i.e. resistance offered by inductor),

Xc = $\frac{1}{\omega c}$ = $\frac{1}{2\pi fC}$

6. Impedance of AC circuit i.e. total resistance offered by AC circuit

Z  =$\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

7. Voltage drop across R, VR = IR

Voltage drop across L, VL = I XL

Voltage drop across C, VC = I Xc

8. Circuit Formula E = IZ (in AC) [E=V=I Req , for r = 0 (in D.C)]

9. Electrical Resonance

At resonance, I is maxm and hence Z is minm as E= IZ

We have, Impedance, Z =$\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

i.e. at resonance, XL = Xc

Resonating frequency, fo  =  $\frac{1}{2\pi }$  $\sqrt{\frac{1}{LC}}$

10. Quality factor = $\frac{{{V}_{L}}\text{ }or\text{ }{{V}_{C}}}{{{V}_{R}}}$

Q. factor =  $\frac{1}{R}$ $\sqrt{\frac{L}{C}}$

11. Average power consumed in AC circuit

Pavg  =  Irms Erms . cos$\phi$

12. Power factor, cos$\phi$ = $\frac{R}{Z}$ = $\frac{R}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}$

13. A.C. Circuit containing inductor with some internal resistance (${{r}_{L}}$)

(i) Impedance of AC circuit, Z  =  $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

(ii) Impedance of AC circuit containing L & R, Z  =$\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

(iii) Total resistance offered by inductor, ZL = $\sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}}$

(iv) Voltage drop across inductor, VL= I ZL

(v) Phase angle ($\phi$), Tan$\phi$ = $\frac{{{Z}_{L}}}{R}$

Also Read: DC Circuit Class 12 Notes