1. Electromechanical Systems
Unit 2
AC Circuits 1 1

2. Aim and Learning Outcomes
Most practical applications in electrical engineering involve alternating current and voltages.
This unit explains
analysis of AC circuits and their operations
use of capacitance transducers.
After completing this unit you should be able to
Analyse passive AC circuits comprising resistors, inductors and capacitors
To determine current flow, voltage distribution and power dissipation.
Identify series resonance.
Analyse power factor improvement circuits.
Describe the operation of capacitance transducers and their use in measuring displacement. 2

3. Content Introduction Resistance Connected to an AC Supply
Inductance Connected to an AC Supply
Capacitance Connected to an AC Supply
Resistance and Inductance in Series with an AC Supply
Resistance and Capacitance in Series with an AC Supply
Resistance, Inductance and Capacitance in Series with an AC Supply
AC Supply in Parallel with Capacitance and with Inductance and Resistance in Series
Power Dissipation
Capacitance Transducers
Problems 3

4. Introduction
Electricity supply systems are normally ac (alternating current).
The supply voltage varies sinusoidal
instantaneous applied voltage, OR
where
Vm = peak applied voltage in volts
f = supply frequency in Hz
t = time in seconds. 4

5. Current and Voltage are in phase
Resistance connected to an AC supply
Instantaneous current, i
Current and Voltage are in phase 5

6. Meters normally indicate rms quantities and this value is
Root Mean Square (rms) Voltage and Current
The “effective” values of voltage and current over the whole cycle
rms voltage is
rms current is
Meters normally indicate rms quantities and this value is
equal to the DC value
Other representations of Voltage or Current are
maximum or peak value
average value
“RMS value of an alternating current is that steady state current (dc) which when flowing through the given resistor for a given amount of time produces the same amount of heat as produced by the alternative current when flowing through the same resistance for the same time” 6

7. Phasor diagram and wave form
Inductance connected to an AC supply
i – instantaneous current i
Current lags Voltage by 90 degree
rms current
Using complex numbers and the j operator
Phasor diagram and wave form
Inductive Reactance 7

8. Phasor diagram and wave form
Capacitance connected to an AC supply i
Current leads Voltage by 90 degrees
rms current
Using complex numbers and the j operator
Capacitance Reactance 8
Phasor diagram and wave form

9. R and L in series with an AC supply
But
and
And
Complex Impedance
Cartesian Form 9
-j indicates that the current lags the voltage

10. phasor diagram constructed with RMS quantities
Complex Impedance:
Cartesian Form:
In Polar Form
phasor diagram constructed with RMS quantities
-L indicates lagging current.
Power factor, p.f.
Complex impedance: 10

11. Exercise:
For the circuit shown below, calculate the rms current I & phase angle L 11
Answer: I = 0.85A -32.10

12. +j signifies that the current leads the voltage.
R and C in series with an AC supply
But
and i
but
Complex Impedance
The current, I in Cartesian form is given by
+j signifies that the current leads the voltage. 12

13. phasor diagram drawn with RMS quantities
Complex Impedance:
I Cartesian form:
In Polar Form
phasor diagram drawn with RMS quantities
+C identifies current leading voltage
Power Factor 13
sinusoidal current leading the voltage

14. 14

15. Exercise:
For the circuit shown, calculate the rms current I & phase angle L 15
Answer: I = 5.32mA 57.90

16. RLC in series with an AC supply
We know that: VC
But
& VL
Complex Impedance VR 16

17. From previous page
The phasor diagram (and hence the waveforms) depend on the relative values of L and 1/C. Three cases must be considered or 17

18. From previous page capacitive resistive inductive Resonant frequency18

19. From previous page
From the above equation for the current it is clear that the magnitude of the current varies with  (and hence frequency, f). This variation is shown in the graph
at o, =
and they may be greater than V
&
fo is called the series resonant frequency.
This phenomenon of series resonance is utilised in radio tuners. 19

20. Exercise: How about you try this ?
For circuit shown in figure, calculate the current and phase angle and power factor when frequency is
(i) Hz, (ii) 1592.Hz and (iii) Hz
How about you try this ?
Answer:
(i) mA o, leading
(ii) 11.04mA, , lagging
(iii) 100mA, 00, 1.0 (in phase) 20

21. AC Supply in Parallel with C, and in Series R &L
Can U name the Laws?
We know that:
Substituting for the different Voltage components gives:
and
Hence, 21

22. How about you try this one too?
Exercise:
For the circuit shown calculate the minimum supply current, Is and the corresponding capacitance C. Frequency is 50 Hz.
How about you try this one too?  22
Answer: ISmin = 3.71A C = 38.6F

23. Power Dissipation net power transfer Therefore, Hence,
We know that:
power dissipation | instantaneous = voltage| instantaneous  current | instantaneous
Hence,
instantaneous voltage,
instantaneous current,
but
&
net power transfer
Therefore, 23

24. Real, Apparent and Reactive PowerIm V P1 Re כ θ i
P = Apparent power
P1 = Real power
P2 = Reactive power P2 P 24

25. Power Factor CorrectionIm V P1 Re O II I i
P22 Pn
P = Apparent power
P1 = Real power
P2 = Reactive power P2 P
P22= New Reactive Power
Pn= New Apparent Power
I= Current to reduce Reactive Power 25

26. Capacitance Transducers
Displacement transducers are often variable capacitors,
Their capacitance varies with movement.
The value may be adjusted by varying either
the distance between the capacitance plates, or
the effective plate area, or
the effective dielectric between the plates
Where
0 = permittivity of free space
r = relative permittivity of dielectric
A = area of overlap between the plates
d = distance between the plates
Capacitance 26

27. To achieve the maximum bridge sensitivity:
To determine the displacement by measuring the capacitance accurately. When the bridge is balanced,
To achieve the maximum bridge sensitivity:
the two capacitors should be equal
the resistances equal to the capacitive reactance at the measuring frequency.
For accurate measurements prevent or minimise:-
stray capacitance between leads and earth
transducer lead inductance
transducer dielectric losses
harmonic distortion (undesired components) in voltage supply 27

28. Linearity of the transducer may be improved by using a
differentially connected displacement device
The transducer is connected to adjacent arms of an ac bridge.
Movement of the central plate increases the capacitance on one side and reduces it on the other. 28

29. Conclusion
AC supply with resistive load, RL in series, RC in series, RLC in series, and RLC in parallel.
Phasor & Cartesian representations.
Phase angle and power factor.
Dissipated Power.
Applications: Capacitance transducer 29

30. Problem Sheet
Q1. A 20V 50Hz supply feeds a 20 Resistor in series with a 100mH inductor. Calculate the circuit (complex) impedance and current.
Q2. A 200V supply feeds a series circuit comprising 250 resistor, 100mH inductor and a 159nF capacitor. Calculate the resonant frequency fo and the corresponding current. Also calculate the current when the frequency is:- fo/3 3fo
Q3. A small company connected to 240V, 50Hz single-phase supply draws a current of 40A at 0.8 power factor lagging. A capacitance is connected across the supply to improve the power factor of the supply current to:
i) unity ii) 0.95 lagging
Calculate the supply current and capacitance in each case.
Q4. The central plate of a differentially connected displacement transducer shown in Fig 2.10c is initially midway between the outer plates. Show that if the central plate is displaced d that the fractional change in the capacitances (C/C) is given approximately by: 30