1. Series and Parallel AC Circuits
SERIES RL CIRCUITS (1)
Circuit above is a series RL network connected to an ac voltage source
Need to find the phasor form of the total impedance of this combination
The total impedance of this series combination is
The magnitude and angle of ZT can be found by converting to polar form:
|ZT| = √[R2+(ωL)2] and θ = tan-1(ωL/R)
The plot of ZT: Im
jωL
|ZT| θ Re R
Series and Parallel AC Circuits

2. Series and Parallel AC Circuits
SERIES RL CIRCUITS (2)
Example: For a series RL combination circuit, R = 300, L = 0.2H and e(t) = 17sin(2000t) V. Find the total equivalent impedance in polar form and rectangular form. Sketch the impedance in the complex plane.
We can use Ohm’s law to find the total current supplied by a voltage source: iT = v / ZT
Mathematical operation must be carried out using phasors since all quantities have both magnitude and angle
The current iT in a series circuit is the same through every series-connected component
The ac voltage drop across each component can be found by multiplying each impedance by the current
v1 = iTZ1, v2 = iTZ2, …
Example: In a series RL circuit, where e(t) = 300 V, R = 200, XL = j100. Find the total current in the circuit. Find the voltage drops across R and L. Verify KVL around the circuit. Draw a phasor diagram showing e, vR, vL and iT. Sketch the voltage waveforms.
Series and Parallel AC Circuits

3. Series and Parallel AC Circuits
SERIES RC CIRCUITS (1)
Above is a series RC network connected to an ac voltage
The total impedance of this combination is
The polar coordinates are
The phasor diagram of the total impedance is Im R Re θ
|ZT|
-j/ωC
Series and Parallel AC Circuits

4. Series and Parallel AC Circuits
SERIES RC CIRCUITS (2)
As in the series RL circuit, the total current supplied to the RC network and the voltage drops can be found by using Ohm’s Law:
iT = e / ZT
vR = iTR and vC = iTXC
Example: For a series RC circuit where e(t) = 18sin(240t + 45) V, R = 3.3k and C = 2.2µF:
a) Find the total current in the circuit in phasor and sinusoidal form
b) Find the voltage drops across the resistor and capacitor in phasor and sinusoidal form
c) Verify KVL around the circuit
d) Draw a phasor diagram showing e, iT, vR and vC
e) Sketch the waveforms of e, vR and vC versus angle
Series and Parallel AC Circuits

5. Series and Parallel AC Circuits
SERIES RLC CIRCUITS (1)
The total impedance of the RLC circuit is
In terms of magnitudes it is: ZT = R + j(|XL| - |XC|)
Inductive and capacitive reactance have opposite signs
Thus net reactance may be either inductive or capacitive, depending which is larger
Polar coordinates are
Series and Parallel AC Circuits

6. Series and Parallel AC Circuits
SERIES RLC CIRCUITS (2) Im Re ZT R θ
jωL
-j/ωC
ωL – 1/ωC
Net reactance
The phasor diagram of the impedance when inductive reactance is greater than the capacitive reactance, i.e. when |XL|>|XC|
The phasor diagram of the impedance when capacitive reactance is greater than the inductive reactance, i.e. when |XC|>|XL| Im
jωL R Re θ
ωL – 1/ωC
Net reactance ZT
-j/ωC
Series and Parallel AC Circuits

7. Series and Parallel AC Circuits
SERIES RLC CIRCUITS (3)
When there is more than one resistor, capacitor and/or inductor in a series circuit, the total impedance has a resistance component equal to the sum of the resistance values and a reactive component equal to the sum of the capacitive reactances subtracted from the sum of the inductive reactances
Example: In a series RLC circuit, e(t) = 100sinωt, R = 800, ZL = j1250 and ZC = -j450.
a) Find the current in polar form
b) Find the voltage drops vR, vL, vC
c) Verify KVL around the circuit
d) Draw a phasor diagram showing e, i, vR, vL, vC
Series and Parallel AC Circuits

8. Series and Parallel AC Circuits
ADMITTANCE
The reciprocal of resistance is conductance, with units Siemens
The reciprocal of impedance is admittance, denoted Y
Y = 1/Z siemens
Impedance is a measure of the extent to which a component impedes the flow of ac current through it
Admittance is a measure of how well it admits the flow of ac current
The greater the admittance, the smaller the impedance, and vice versa
Resistance R is one form of impedance Z, conductance G is one form of admittance Y
Phasor form of conductance is:
Series and Parallel AC Circuits

9. Series and Parallel AC Circuits
SUSCEPTANCE
Reactance is another special case of impedance
The reciprocal of reactance is called susceptance, B
B = 1/X siemens
There are two types of susceptance
Inductive susceptance:
Capacitive susceptance:
Z = 1/Y; R = 1/G; XL = 1/BL; XC = 1/BC
Example: Find the admittance of a 5 resistor; a 5mH inductor at f = 60Hz; a 0.2µF capacitor at ω = 1.25106 rad/s.
Example: Find the admittance Y corresponding to Z = 30 + j40 . Draw a phasor diagram showing Y in the complex plane.
Series and Parallel AC Circuits

10. PARALLEL AC CIRCUITS (1)
Above shows a parallel connected set of impedances to an ac source
Total admittance of the circuit is the sum of the admittances of the parallel connected components, i.e.
YT = Y1 + Y2 +… + Yn = 1/Z1 + 1/Z2 +…+ 1/Zn
Example: Find the total admittance of a 20mH inductor, a 1k resistor and a 0.16µF capacitor connected in parallel to an ac voltage source of 15sin(25103t) V. Draw a phasor diagram showing the total admittance in the complex plane. What is the total admittance if the frequency of the voltage source is doubled?
Series and Parallel AC Circuits

11. PARALLEL AC CIRCUITS (2)
The total impedance of a network is the reciprocal of its total admittance, ZT = 1/YT
For a parallel network:
Example: Three components (an inductor with impedance j10, a resistor with impedance 2, and another inductor with impedance j5) are connected in parallel. Find the total equivalent impedance of this network.
Example: A capacitor with impedance –j100 and an inductor with impedance j500 are connected in parallel. Find this network’s total equivalent impedance.
Series and Parallel AC Circuits

12. PARALLEL AC CIRCUITS (3)
The same voltage appears across every parallel-connected impedance
So current can be found using Ohm’s law and KCL
Series and Parallel AC Circuits

13. PARALLEL AC CIRCUITS (4)
Example: For the circuit below, find the current in each impedance. Show that iT = eYT. Draw a phasor diagram showing e, iT, i1, i2 and i3.
Series and Parallel AC Circuits

14. Series and Parallel AC Circuits
AC CURRENT SOURCES (1)
Like a dc current source, an ideal ac current source supplies the same constant current to whatever network is connected across its terminals
For ac, the current is constant in the sense that its peak value does not change
The same symbol is used for an ac current source as for a dc current source
The direction of the arrow is just a phase reference, since current reverses direction every half cycle
Reversing the arrow is the same as multiplying the current by -1, which is the same as adding or subtracting 180 from its phase angle
Series and Parallel AC Circuits

15. Series and Parallel AC Circuits
AC CURRENT SOURCES (2)
Example: For the circuit below, find the voltage vT across the ac current source. Find the currents i1 and i2. Show that i = i1 + i2. Draw a phasor diagram showing i, vT, i1 and i2. Sketch the waveforms of i, i1 and i2 versus angle.
Series and Parallel AC Circuits

16. POWER IN CIRCUITS CONTAINING REACTANCE (1)
Recall that the average power dissipated by a resistance carrying sinusoidal ac current can be found by
Above can be used to find Pavg dissipated by a resistor in a circuit containing reactance, provided the values used in the computations are indeed those of the voltage across and/or current through the resistor itself
In many practical circuits, it is sometimes necessary to compute Pavg when the resistance or resistor voltage/current are not known
Series and Parallel AC Circuits

17. POWER IN CIRCUITS CONTAINING REACTANCE (2)
Above is an ac circuit with a general impedance Z
Assuming that the voltage supplied has angle 
Angle between voltage and current is  - ( - ) = 
The angle between the voltage applied to a network and the total current supplied to it equals the angle of the impedance of the network
Resistive component of impedance Z is R = |Z|cos ohms
Since only resistance in network dissipates power
Series and Parallel AC Circuits

18. Series and Parallel AC Circuits
POWER FACTOR
Equations on previous slide provide a means for computing the average power in terms of voltage e across whole network and total supply current i
Cos is called the power factor
For a purely resistive network, voltage and current are in phase, so power factor = cos0 = 1 → Pavg = EpIp / 2
For a purely reactive network, voltage and current are separated by ±90, and cos ±90 = 0 → Pavg = 0, i.e no power is dissipated by purely reactive components (capacitors and inductors)
Since Vp = 2Veff and Ip = 2Ieff we can derive
Series and Parallel AC Circuits

19. EXAMPLE OF POWER IN AC CIRCUITS
In the circuit above:
Find the power factor
Use the power factor to find the average power dissipated in the network
Verify that the power computed above is the same as the power computed by using the voltage across and current through the resistor
Series and Parallel AC Circuits