Module 2.  A.C. series and parallel circuits

 

Lesson 6

A.C. SERIES AND PARALLEL CIRCUITS

6.1  A.C. Circuit

When an alternating current flows in closed loop or a path, it is called an a.c. circuit. Different elements of an a.c. circuit may be any or in combination of following:

1.      Resistance: An electrical element which causes opposition to the passage of an electric current through that element.

2.      Inductance: In electromagnetism and electronics, inductance in the circuit "induces" (creates) a voltage (electromotive force) in both the circuit itself (self-inductance) and any nearby circuits (mutual inductance). Inductance is typified by the behaviour of a coil of wire in resisting any change of electric current through the coil.

3.      Capacitance: Capacitance is the ability of a body to store an electrical charge. Any element or structure that is capable of being charged, either with static electricity or by an electric current exhibits capacitance.

If the voltage applied to an a.c. circuit in sinusoidal, the resulting alternating current is sinusoidal. Also the frequency of the alternating current will be equal to that of applied voltage. The opposition to the flow of current in an a.c. circuit may be due to:

1.      Resistance R

2.      Inductive reactance (XL = ωL)

3.      Capacitive reactance (XC = )

Different types of A.C. Circuit can be listed as follows:

1.      A.C. Circuit with only one element

a)      Resistance

b)      Inductance

c)      Capacitance

2.  A.C. Series Circuit

a)      R-L Series Circuit

b)      R-C Series Circuit

c)      R-L-C Series Circuit

3. A.C. Parallel Circuit

6.2  A.C. Circuit with Only One Element

6.2.1  A.C. Circuit with only one element-resistance

The figure 6.1 shows an a.c. circuit with a pure resistance of R Ω

Fig. 6.1 A.C. Circuit with only one element-Resistance

The current in the circuit is i = v/R ......... (1)

The instantaneous value of alternating voltage is given by:

v = VmSin ωt ......... (2)

From equation (1) and (2)     

......... (3)
The value of current will be maximum when Sin
𝜔t=1        

         ......... (4)     

From equation (3) and (4)

i = ImSin ωt ......... (5)

In an a.c. circuit having resistance as any element, the phase difference between voltage and current is zero. In other words it can be said that current is in phase with the voltage power.           

We know instantaneous power p = vi

 From equation (2) and (5) we have

Considering average power           

                            


 

P = Vrms Irms

p = v i where V = rms voltage, I = rms current

6.2.2  A.C. Circuit with only one element-inductance

Figure 6.2 shows an a.c. circuit with pure inductance of L Henry

Fig. 6.2 A.C. Circuit with only one element-Inductance

The sinusoidal voltage can be given as

The e.m.f in the coil due to the current I flowing in the circuit can be given as            

......... (2)

The induced e.m.f is also called e.m.f opposes the change of current in the coil .Back e.m.f. induced in the coil is equal and opposite of the applied voltage. Thus equation     

Integrating



......... (3)

 

i will be maximum when value of ) is one

  ......... (4)

      

In the above equation

ωL  is also known as inductive reactance XL of the coil.

  ω =angular velocity

  f =frequency in hertz

  L =inductance in henry

  XL=inductive reactance in Ω (ohms)

Note: XL is the opposition offered by pure inductance to the flow of an alternating current

From equation (3) and (4)       we have

......... (5)

From equation (1) and (5) it can be seen that current in a inductive circuit lags behind the voltage by radians or 90o.It is well represented in the phasor diagram shown in fig. 6.3

 

Fig. 6.3 Phasor diagram for current and voltage in an inductive circuit

Power

Instantaneous power p = v i

                                                                                                   

Average power consumed over one cycle 


This shows that power absorbed in a circuit having only inductance element is zero

6.2.3  A.C. circuit with only one element- capacitance 

Fig. 6.4 Shows a circuit with capacitance C farads

                   

Fig. 6.4 Circuit with capacitance C farads

The value of alternating voltage is given as                        

The charge on the capacitor q=c v

The current in the circuit is  

          

......... (1)

The value of the current will be maximum when                            

From equation (1)                              

This equation shows that in a circuit with capacitance as an only element, the current leads the voltage by 90o (Fig. 6.5)

 

Fig. 6.5 Current leads the voltage by 90o in a circuit with capacitance

Power

                  Instantaneous power is given by

p = v i

         

The average power over one complete cycle is p = Zero                                     

The power absorbed in a circuit with pure capacitance is zero.

6.3  A.C. Series Circuit

There are three major types of circuit as follows

1.      R-L Series Circuit

2.      R-C Series Circuit

3.      R-L-C Series circuit

6.3.1  R-L series circuit

Figure 6.6 shows a pure resistance and inductance connected in series

Fig. 6.6 A.C. circuit with series resistance and inductance

Fig. 6.7 Phasor diagram for a.c. series circuit

From the phasor diagram              

V2 = (VR)2 + (VL)2

      

    

  Here,

        V = r.m.s value of applied voltage

         I = r.m.s. value of current

         VR = voltage drop across R = IR

         VL = voltage drop across L = IXL

             Z = Impedance of the circuit and it is measured in Ω ohms

               
                        

 Here ϕ is known as phase angle

Voltage leads current by ϕ angle. In other words it can be said that current lags voltage by ϕ angle. Figure 6.8 shows the impedance triangle

Fig. 6.8 Impedence triangle for a.c. series circuit

Impedance Z

Here, Z = Impedance and R = Resistance

          XL=Inductive reactance

            ϕ = Phase angle

           Cos ϕ = power factor of the circuit

Power   P= VI Cos ϕ

                = (Z.I) ICos ϕ

              P = I2R

In a series R-L Circuit power is consumed in resistance only. Inductance consumes zero power. The unit of power is watt.                         

6.3.2  R-C series circuit

Fig. 6.9 shows a pure resistance and capacitance connected in series.

Fig. 6.9 A.C. circuit with series resistance and capacitance

Fig. 6.10 Phasor diagram for a.c. series resistance and capacitance circuit

From the phasor diagram figure 6.10

V2 = (VR)2 + (-VC)2

      

     

  Here,

        V = r.m.s value of applied voltage

         I = r.m.s. value of current

         VR = Voltage drop across R = IR

         VC = Voltage drop across C = IXC

             Z = Impedance of the circuit and it is measured in ohms



Here ϕ is known as phase angle.

Current leads voltage by ϕ angle. In other word can be said that voltage lags current by ϕ angle. The figure 6.11 shows the impedance triangle.

Fig. 6.11 Impedance triangle for a.c. series circuit

Impedance Z

                                                                                                         

Here, Z = Impedance and R=Resistance

          XC = Capacitive reactance

          = Power factor of circuit

           C = Capacitance

           XC = = Capacitive reactance

Power   P = VICosϕ

Here,

         V = r.m.s value of applied voltage

          I = r.m.s. value of current

         = Power factor of circuit

6.3.3  R-L-C series circuit

When a pure resistance R ohms, pure inductance L Henry and pure capacitor of capacitance C farad are connected in series it is known as R-L-C Series Circuit.

Fig. 6.12 A.C. series circuit for resistance, inductance and capacitance

 

Fig. 6.13 Phasor diagram for resistance, inductance and capacitance series circuit                                             

The voltage drop across each element is given as:

a)      Resistance (R) = VR = IR (in phase with current I)

b)      Inductance (L) = VL = IXL (leads current by 90o)

c)      Capacitance (C) = VC = I.XC( lags current by 90o)

In the phasor diagram VL is leading current I by 90o and VC is lagging current by 90o. So it is evident that VL and VC are at 180o to each other. In technical terms it is said to be 180o out of phase with each other. The circuit will behave like inductive or capacitive manner depending upon voltage drop VL or VC w.r.t current I.

From the phasor diagram:

                           

                                   

                                      

                        

Where

          Z = Impedance of the circuit which offers opposition to current flow

Phase Angle

Again from the phasor diagram



Three cases of R-L-C Series Circuit

          The equation for impedance is given as:

          

Case 1: When XL>XC

The term (XL-XC) is positive.

The circuit works as an R-L Series Circuit.

Current lags behind voltage.

Phase angle is positive.

Power angle is positive.

Power factor is lagging.

Current flowing in circuit i,

Case 2: When XC>XL

The term (XL-XC) is negative.

The circuit works as an R-C Series Circuit.

Current leads over voltage.

Phase angle is negative.

Power factor is negative.

The current flowing in the circuit i,

Case 3: When XL = XC

       The term (XL-XC) = 0.

       The circuit works as pure resistance.

       Current is in phase with voltage.

       Phase angle is Zero.

       Power factor = 1.

       The current flowing in the circuit i,

6.4  True Power and Reactive Power

Table 6.1 Power in the electrical circuit

 

 

Formula

Units

1.      True power

 

 

 

 

 

 

 

 

 

         Power consumed by watt meter

         Is the useful work and the current is in phase with the voltage

         Is the power consumed by resistance

         It is also known as active power

 

PTrue = VI Cosϕ

 

 

 

 

 

 

 

 

 

Watts

 

 

 

 

 

 

 

 

 

 

2.      Reactive power

 

 

 

 

 

 

 

 

         Power consumed in L or C in a circuit is zero but the circulating power is termed as reactive power

         Does no useful work and current is 90o out of phase with voltage.

         Reactive power cannot be measured by wattmeter.

PReac = VI Sinϕ

 

 

 

 

 

 

 

 

 

 

VAR

 

 

 

 

 

 

 

 

 

 

3.      Apparent power

         It is defined as the product of Voltage and Current

Papparent = VI

 

VA

 

6.5  Power Triangle

Figure 6.14 shows the power triangle for an A.C. circuit.

Fig. 6.14 Power triangle for an A.C. circuit

True power PTrue = VI Cos ϕ

Reactive Power PReac = VI Sin ϕ

Apparent power Papp = VI

From power triangle           

6.6  A.C. Parallel Circuits

In a.c. circuits R, L and C are connected in parallel. Voltage across each element is same but the current flowing through it is different. Equipments, lights and circuits are connected and operated in parallel.  Parallel connection gives advantage that each equipment, appliance or device can be operated independently having separate switches for on/off. Parallel circuits are analysed using following methods:

1.      Phasor diagram

2.      Admittance method

3.      Symbolic methods

Numerical

1. A coil of R = 100 ohm and L = 125 milli Henry is connected across  alternating voltage e = 250 sin 100𝛑t. Determine:

a.      Impedance

b.      Current through the coil

Solution:

2πft = 100πt

f = 50 Hz

Inductive reactance XL =  

                                    = 2πfL

                                    = 2π50x125x 10-3

                                    = 39.25 ohm

Impedance Z     

 

                    Z   

Z = 107. 42 ohm

 

Current through coil I = em/Z

                                     = 250/107.42 = 2.32 A

2. A resistance of 50 ohm and capacitor of 200 F are connected across  220 V, 50 Hz voltage supply. Determine:

a.      Impedance

b.      Current through the coil

c.       Power factor

Capacitive reactance = XC =  =  =  = 15.92 ohm

 

Impedance Z     

 

Z   

Z = 52. 47 ohm

Current I = V/Z = 220/52.47 = 4.19 A

Power factor Cos ϕ = R/Z = 50/52.47 = 0.953 leading

3. A resistance of 50 ohm, inductance 75 mH and capacitor of 25 F are connected across  220 V, 50 Hz voltage supply. Determine:

a.      Impedance

b.      Current through the coil

c.       Power factor

Inductive reactance XL =  

                                    = 2πfL

                                    = 2πx50x75x 10-3

                                     = 23.55 ohm

 Capacitive reactance = XC =  =  =  = 127.38 ohm

  Impedance  Z   

Z

Z = 115.24 ohm

  Current I = V/Z = 220/115.24 = 1.9 A

  Power factor Cos ϕ = R/Z = 50/115.24 = 0.433 leading