Alternating Current

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Basic Concepts of A.C   Physicsminds.blogspot.com
A waveform is a representation of how alternating current varies with time. The most
familiar AC waveform is the sine wave, which derives its name from the fact that the
current or voltage varies with the sine of the elapsed time. Other common AC waveforms
are the square wave, the ramp, the sawtooth wave, and the triangular wave. Their general
shapes are shown below.

various types of output waveforms of ac sources

The principles you have learned for direct current, such as Ohm’s law, Kirchhoff’s laws
and the principles used for resistors in series and parallel are also applicable to alternating
current.

The Sinusoidal AC Waveform

The most common AC waveform is a sine (or sinusoidal) waveform.

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The vertical axis represents the amplitude of the AC current or voltage, in
amperes or volts.
The horizontal axis represents the angular displacement of the waveform. The
units can be degrees or radians.

A complete change of current or p.d. from a particular value and back to the same
value in the same direction is known as a cycle.
The time taken for a complete cycle is known as the period T of the alternating
current or p.d.
The number of complete cycles in one second is known as the frequency of the
alternating current or p.d. In most countries, the frequency of the alternating
current is 50 Hz.
Peak amplitude is the height of an AC waveform as measured from the zero
mark to the highest positive or lowest negative point on a graph. Also known as
the crest amplitude of a wave.

Peak-to-peak amplitude is the total height of an AC waveform as measured from
maximum positive to maximum negative peaks on a graph. Often abbreviated as
“P-P”.

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Phasor Diagram
To account for the different phases of the voltage drops, vector techniques are
used
Represent the voltage across each element as a rotating vector, called a phasor
The diagram is called a phasor diagram
Phasor diagram-to visualize the phase of alternating quantities I and V.
A phasor is a rotating vector used to represent a quantity that varies sinusoidally
with time.

Resistors in an ac circuit

   An AC circuit consists of a combination of circuit elements and an AC generator
or source
   Consider a circuit consisting of an AC source and a resistor.

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The sinusoidal current in the circuit can be represented by the equation
I  I 0 sin 2ft
The potential difference across the resistor of resistance R is
V  IR  ( I 0 sin 2ft ) R  ( I 0 R) sin 2ft  V0 sin 2ft

V  V0 sin 2ft

    The graph shows the variation of the current I and the potential difference V with
time t.

Graph (b)       shows variation of I and V with time
 Both achieve max at same time =in phase with each other

Phasor diagram (c) -to visualize the phase of an alternating quantities I and V.
I and V are in phase, hence their phasors are drawn parallel to each other and they rotate
about the point O with the same frequency f.

The power dissipated in a resistor of resistance R when a current I flows is given by,
P  I 2R
For a complete cycle, mean Power = mean value of I2R = (mean value of I2) x R
mean P  I rms  R
2

where I rms  mean value of I 2
2

Hence   I rms  mean valueof I 2           Irms=root mean square current of the ac

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2
I0   I
I rms          0
2      2
V02 V0
Vrms       
2   2
AC ammeters and voltmeters are designed to read rms values

Example
An ac voltmeter, when placed across a 12  resistor, reads 117 V. What are the
maximum values for the voltage and current?

Solution:

Vmax  2Veff  1.414(117 V);        Vmax = 165 V

 117 V 
I max  2 I eff  1.414         ; Imax = 13.8 A
 12  

Example
An ac current of rms value 4.0 A and frequency 50 Hz flows in a circuit containing a
resistor. What is the (a) peak current and (b) the value of the current 6 x 10-4 s after it
changes direction?

Answer: (a) Using I 0  2I rms  I0 =5.66A

(b) Using   I  I 0 sin 2ft    I=1.06 A

Capacitors in an ac circuit
Pure capacitor – no resistor & no self induction

o       The current starts out at a large value and charges the plates of the capacitor
 There is initially no resistance to hinder the flow of the current
while the plates are not charged

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o      As the charge on the plates increases, the voltage across the plates increases and
the current flowing in the circuit decreases
o      The current reverses direction
o      The voltage across the plates decreases as the plates lose the charge they had
accumulated
o      The voltage across the capacitor lags behind the current by 90°
o      The impeding effect of a capacitor on the current in an AC circuit is called the
capacitive reactance and is given by:

Suppose that the alternating p.d. applied to the capacitor is given by

V  V0 sin 2ft
On capacitor, Q  CV  CV0 sin 2ft
d           
 CV0 sin 2ft   CV0  sin 2ft   CV0 (2ft ) cos(2ft )
dQ d
From I 
dt dt                         dt         
The current in the circuit

 I  I 0 sin(2ft               )
2
where I 0  CV0 (2f ) is the peak current.

Compare (6.7) and (6.8): V and I  not in phase

Phase difference between V and I  radians
2
                                     
2                                     2

From graph (a)          Imax at t = 0 and Vmax at t = T/4

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
 phase diff =
2
From phasor diagram (b)  when both the phasors are rotated in the direction shown at

2

From I0 =CV0 ( 2f )
V0      1
       where V0 =            2 Vrms and I0 =   2 Irms
I 0 2fC
V0       2Vrms         Vrms    1
                             X c  reactance of the capacitor
I0       2 I rms       I rms 2fC

1
Xc 
2fC

Xc

frequency

Variation of Xc with frequency
From Ohm’s law, V =IR

Ohm’s Law for a capacitor in an AC circuit

     Vc  I rms X c                                  …(6.10)

Example

A 6-F capacitor is connected to a 40-V, 60 Hz ac line.
(a) What is the reactance?
(b) What is the effective ac current in the circuit containing pure capacitance?

Solution:
1            1
(a) X C                                ;           XC = 442 
2 fC 2 (60 Hz)(6 x 10-6 F)

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Veff        40 V
(b) I eff                    ;   Ieff = 90.5 mA
XC         442 

Inductors in an ac circuit
o Consider an AC circuit with a source and an inductor
o The current in the circuit is impeded by the back emf of the inductor

Current flow in an inductor:        0I  I sin 2ft
When a charging current flows in the inductor, a back emf is produced.

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dI
 back   L
dt
Applying Kirchoff’s 2nd Law to the circuit,
dI
 back  L  0
dt
dI      d
V L         L ( I 0 sin 2ft)  2fLI0 cos(2ft)
dt      dt
      
V  V0 sin  2ft  
      2
Where the peak voltage, V0  2fLI 0

Compare (6.11) and (6.12);    I  I 0 sin 2ft
       
V  V0 sin  2ft  
       2

2

V0 Vrms
From (6.13); V0  2fLI 0  X L               2fL
I 0 I rms

V

f


2

The effective resistance of a coil in an AC circuit is called its inductive reactance and is
given by XL

Ohm’s law for inductor:   Vrms  I rms X L

Basic Concepts of A.C                                  Physicsminds.blogspot.com
Example

A 50-mH inductor of negligible resistance is connected to a 120-V, 60-Hz ac line.
(a) What is the inductive reactance?
(b) What is the effective ac current in the circuit?

Solution:
(a) X L  2 fL  2 (60 Hz)(0.050 H)  18.85 ; ;             XL = 18.9 

Veff        120 V
(b) I eff                      ;       Ieff = 6.37 A
XL         18.85 

Resistors and Capacitors in Series

R & C in series  the same I flows in R & C
Across R  VR=IR  VR and I in-phase

Across C  VC=IXC  VC lags behind I by              rad
2

 VC lags behind VR by    rad
2

Resultant V across R & C, V 2  V R2  VC2  I 2 ( R 2  X C )
2

1
V  I R2 
2fC2
V           1
 R2           Z = impedance of the circuit
I        2fC2

1
Z  R2                             Unit:  (ohm)
2fC2

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Figure 16 shows the variation of the impedance Z with frequency. For a very high
frequency (f   ), the impedance Z equals the resistance R of the resistor.

Phase difference between I & V :
VC IX C     1
tan            
VR   IR   2fCR

Example
An ac current of rms value 1.5 mA and angular frequency  =100 rad/s flows through a
resistor of 10 k  and a 0.50  F capacitor joined in series. Calculate
(a)     the rms voltage(potential difference) across the resistor
(b)     the rms voltage across the capacitor
(c)     the total rms voltage across the resistor and the capacitor
(d)     the rms voltage across the resistor when the current is maximum
(e)     the impedance in the circuit

Solution
(a)     VR =IR=15 V
(b)     VC = IXC =30 V
(c)     Using V 2  V R2  VC2  Vtotal = 33.5 V
(d)     Across resistor, V and I in phase, where as across the capacitor, V lags behind

2
Therefore, when the current is maximum, the voltage across the capacitor is
zero and the voltage across the resistor is maximum. Hence the voltage across
the resistor when the current is maximum
V=I0R= 2I rms =21.2 V
V
(e)         Impedance, Z  rms  2.23  10 4 
I rms

Basic Concepts of A.C                             Physicsminds.blogspot.com
6.6 Resistors and Inductors in series

Across R  VR=IR           VR and I in-phase

2

2

Resultant V across R & L,   V 2  VR2  VL2  I 2 ( R 2  X L )
2

    V  I R 2  (2fL) 2
V
Impedance of the circuit   Z      R 2  (2fL) 2
I

V   IX   2fL
Phase angle between I & V : tan   L  L 
VR   IR   R

Basic Concepts of A.C                          Physicsminds.blogspot.com
Example
A solenoid has inductance 2.5 H and resistance 6.0  .
(a) Sketch graphs to show how the resistance R and the reactance X of the solenoid varies
with the frequency.
(b) At what frequency is the resistance of the solenoid equal to 1% of its reactance?

Solution:
(a)

XL

R=constant

frequency

(b)    When R=1% of reactance
1
6=      2  f  2.5  f=38.2 Hz
100

Basic Concepts of A.C                             Physicsminds.blogspot.com
Resistor, Capacitors and Inductors (RCL) in series

Figure 18

When a resistor of resistance R, a capacitor of capacitance C and an inductor of
inductance L are connected in series with an ac supply, the potential difference
across R, C and L are, respectively,
Across R  VR=IR  VR and I in-phase

Across C  VC=IXC  VC lags behind I by         rad
2

2
The magnitude of VL can be greater or less than VC.

   Phasor Diagram for RLC Series Circuit:

o The voltage across the resistor is on the +x axis since it is in phase with
the current
o The voltage across the inductor is on the +y since it leads the current by
90°

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o The voltage across the capacitor is on the –y axis since it lags behind the
current by 90°

Magnitude VL can be greater or less than VC

1 2
Resultant V across R, C & L, V 2  VR2  (VL  VC ) 2  I 2 [ R 2  (2fL         )
2fC
V                   1 2
Impedance of the circuit        Z       R 2  (2fL       )
I                 2fC

(VL  VC ) 2fL  2fC
1
Phase angle between I & V : tan                         …
VR            R

Ohm’s Law can be applied to the impedance

                  ΔVmax = Imax Z

6.8 Resonance
In the RLC circuit, if XL = XC then the impedance becomes Z=R.
1
Since  X L  2fL and X C            , there would be a certain frequency known as the
2fC
resonance frequency f0 where the value of XL = XC.

When this happens, the impedance Z will have the smallest value and the current will be
maximum.

The resonance frequency can be determined as follows:
1
When XL = XC  2f 0 L            4 2 f 02 LC  1
2f 0 C
1
f0 
2 LC
 resonance frequency of the RLC series.

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Figure 19

Figure 19 shows the variation of the impedance with the frequency f and (b) shows the
1
variation of the current I with the frequency. When f 0            , the impedance is
2 LC
least and the current is maximum. The frequency f0 is known as the resonance frequency
of the series RLC circuit.

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Current amplitude vs angular frequency for series RLC circuit with various R values.

Example

A 300-  resistor, a 3-F capacitor, and a 4-H inductor are connected in series with a 90-
V, 50 Hz ac source. What is the net reactance of the circuit? What is the impedance?
What is the resonant frequency?

Solution:
1
XC                            1061 ;    X L  2 (50 Hz)(4 H)  1257 ;
2 (50 Hz)(3 x 10-6 F)

Net Reactance:        XL – XC = 196 

Z  R 2  ( X L  X C )2  (300 )2  (1257   1061 ) 2 ;   Z = 358 

1           1
fr                              ;         fr = 45.9 Hz
2 LC 2 (4 H)(3 x 10-6 F)

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Power in an ac circuit

o No power losses are associated with capacitors and pure inductors in an AC
circuit
o In a capacitor, during one-half of a cycle energy is stored and during the other half
the energy is returned to the circuit
o In an inductor, the source does work against the back emf of the inductor and
energy is stored in the inductor, but when the current begins to decrease in the
circuit, the energy is returned to the circuit
o The power in any component in an electric circuit is given by:P=IV

From   I  I 0 sin 2ft
V  V0 sin 2ft

P  IV  I 0V0 sin 2 (2ft )

o Mean power, P = mean value of        I 0V0 sin 2 ( 2ft )
2
1                    Vrms
P  I 0V0  I rms  R 
2

2                     R
In a resistor, the power dissipated in the form of heat.

Basic Concepts of A.C                               Physicsminds.blogspot.com
Example
An inductor dissipates heat at a rate of 10 W when an alternating current of 0.5 A flows
in it. If the reactance of the inductor at the frequency of the supply is 30  , what is its
impedance?

Solution:
Since heat is dissipated from the inductor, the inductor is not a pure inductor and has a
resistance R. Power is dissipated due to this resistance.
Using P  I R  R= 40 
2

Impedance Z  R 2  X L  50
2

Example
A series ac circuit consists of a 100-  resistor, a 0.2-H inductor, and a 3-F capacitor
connected to a 110-V, 60 Hz ac source. What are the inductive reactance, the capacitive
reactance and the impedance for the circuit?

Solution:
1
XC                            884  ;   X L  2 (60 Hz)(0.2 H)  75.4 
2 (60 Hz)(3 x 10-6 F)

Z  R 2  ( X L  X C )2  (100 )2  (75.4   884 ) 2 ;   Z = 815 

Example

When a 6-  resistor and a pure inductor are connected to a 100-V, 60 Hz ac line, the
effective current is 10 A. What is the inductance? What is the power loss through the
resistor and what is the power loss through the inductor?

Solution:
110 V
Z         11.0 ; X L  Z 2  R 2  (11.0) 2  (6 ) 2 ;
10 A

XL   (9.22 )
XL = 9.22  L                   ; L = 24.5 mH
2 f 2 (60 Hz)

P = I2R = (10 A)2 (6.00  );     P = 600 W;

No Power lost in inductor: 0

Basic Concepts of A.C                               Physicsminds.blogspot.com
Basic Concepts of A.C   Physicsminds.blogspot.com

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