ELECTRICITY & MAGNETISM (Fall 2011)

1. ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 12 BY MOEEN GHIYAS

2. (Series Circuit – Chapter 5) Introductory Circuit Analysis by Boylested (10th Edition) TODAY’S lesson

3. Today’s Lesson Contents • Notation • Internal Resistance of Voltage Sources • Voltage Regulation • Measurement Techniques • Applications • Solution to Problems

4. Notation – Voltage Sources and Ground • Three ways to sketch the same series dc circuit • If two grounds exist in a circuit and no connection is shown between them, even then such a connection exists for the continuous flow of charge.

5. Notation – Voltage Sources and Ground • On large schematics where space is at a premium and clarity is important, voltage sources may be indicated as in fig (a) rather than as illustrated in fig (b)

6. Notation – Voltage Sources and Ground • On large schematics where space is at a premium and clarity is important, voltage sources may be indicated as fig (a) rather than as illustrated in fig (b)

7. Notation – Voltage Sources and Ground • In schematics, the potential levels may also be indicated to permit a rapid check of the potential levels at various points in a network with respect to ground to ensure that the system is operating properly

8. Notation – Double-Subscript Notation • Voltage is an across variable and exists between two points resulting in a double-subscript notation • In fig, since a is the first subscript for Vab, point a must have a higher potential than point b if Vab = +ve value. • If point b is at a higher potential than point a, then Vab = -ve value.

9. Notation – Single-Subscript Notation • The single-subscript notation Va specifies the voltage at point a with respect to ground (zero volts). Thus for voltage at point b w.r.t to ground, we have Vb • If the voltage is less than zero volts, a negative sign must be associated with the magnitude of Va

10. Notation – General Comments • Also the voltage Vab can be determined using Eq. Vab = Va – Vb • For fig below:

11. Notation • Example – Find the voltage Vab for the conditions of fig • Solution: • Note the negative sign to reflect the fact that point b is at a higher potential than point a.

12. Notation • Example – Find voltage Va for the configuration of Fig • Solution:

13. Notation • Example – Find the voltage Vab for the configuration. • Solution:

14. Notation & Voltage Divider Rule • Example – Using the voltage divider rule, determine the voltages V1 and V2 of fig. • Solution: Circuit Redrawn, • . From • . voltage divider rule,

15. Notation & Voltage Divider Rule • Example – For the network of fig • Calculate Vab. • Determine Vb. • Calculate Vc.

16. Notation & Voltage Divider Rule • Calculate Vab. Solution: • Determine Vc. Solution:

17. Notation & Voltage Divider Rule • Determine Vb. • Solution: • or

18. Internal Resistance of Voltage Sources • Every voltage source, whether a generator, battery, or laboratory supply (fig (a)) has some internal resistance. • The equivalent circuit of any voltage source appears as shown in fig (b).

19. Internal Resistance of Voltage Sources • The effect of the internal resistance on the output voltage is important to study in order to understand unexpected changes in terminal characteristics of voltage source.

20. Internal Resistance of Voltage Sources • The ideal voltage source has no internal resistance and an output voltage of E volts with no load or full load as shown in fig (a).

21. Internal Resistance of Voltage Sources • In the practical case [fig(b)], where we consider the effects of the internal resistance, the output voltage will be E volts only when no-load (IL = 0) conditions exist.

22. Internal Resistance of Voltage Sources • When a load is connected [fig (c)], the output voltage of the voltage source will decrease due to the voltage drop across the internal resistance.

23. Internal Resistance of Voltage Sources • By applying Kirchhoff’s voltage law around the indicated loop of fig (c), we obtain • Since • We have • And we get an important relation • If Rint is not known, it can be found

24. Internal Resistance of Voltage Sources • A direct consequence of the loss in output voltage is a loss in power delivered to the load. • Multiplying both sides of eq. by the current IL in the circuit, we obtain

25. Internal Resistance of Voltage Sources • For a dc generator, a plot of the output voltage versus current appears in fig

26. Internal Resistance of Voltage Sources • Note that any increase in load demand causes a drop in terminal voltage due to the increasing loss in potential across the internal resistance.

27. Internal Resistance of Voltage Sources • At maximum current IFL, the voltage across the internal resistance is Vint = IFLRint = (10 A)(2 ) = 20 V, and the terminal voltage has dropped to 100 V—a significant difference (from 120V) even if you stay below the listed full-load current.

28. Internal Resistance of Voltage Sources • Eventually, if the load current were permitted to increase without limit, the voltage across the internal resistance would equal the supply voltage, and the terminal voltage would be zero.

29. Internal Resistance of Voltage Sources • The larger the internal resistance, the steeper is the slope of the characteristics of fig • In fact, for any chosen interval of voltage or current, the magnitude of the internal resistance is given by

30. Internal Resistance of Voltage Sources • Example – Before a load is applied, the terminal voltage of the power supply of is set to 40V. When a load of 500Ω is attached, the terminal voltage drops to 38.5 V. What happened to the remainder of the no-load voltage, and what is the internal resistance of the source?

31. Internal Resistance of Voltage Sources • Solution: • The difference of 40 V – 38.5 V = 1.5 V now appears across the internal resistance of the source. • The load current IL = 38.5 V/0.5 kΩ = 77 mA. • . Applying Eq. 38.5V

32. Internal Resistance of Voltage Sources • Example – The battery of fig has an internal resistance of 2Ω . Find the voltage VL and the power lost to the internal resistance if the applied load is a 13Ω resistor. • Solution:

33. Voltage Regulation • If a supply is set for 12 V, it is desirable that it maintain this terminal voltage, even though the current demand on the supply may vary. • A measure of how close a supply will come to ideal conditions is given by the voltage regulation characteristic.

34. Voltage Regulation • By definition, the voltage regulation (VR) of a supply between the limits of full-load and no-load conditions (Fig. 5.56) is given by the following:

35. Voltage Regulation • We see for ideal conditions, VFL = VNL and VR% = 0. • Therefore, the smaller the voltage regulation, the less the variation in terminal voltage with change in load. • It can be shown with a short derivation that the voltage regulation is also given by • The smaller the internal resistance for the same load, the smaller the regulation and more ideal the output.

36. Voltage Regulation • Example - Calculate the voltage regulation of a supply having the characteristics of Fig. 5.53. • Solution:

37. Voltage Regulation • Example - Determine the voltage regulation of the supply of fig with internal resistance of 19.48Ω and load resistance as 500Ω. • Solution: 38.5V

38. Measurement Techniques - Ammeters • Ammeters are placed in series with the branch in which the current is to be measured • For minimal impact on the network behaviour, ammeter’s resistance should be very small (ideally zero ohms) compared to the other series elements of the branch

39. Measurement Techniques - Ammeters • If the meter resistance approaches or exceeds 10% of branch resistance R, it will have a significant impact on the current level it is measuring. • It is also noteworthy that the resistances of the separate current scales of the same meter are usually not the same. • In fact, the meter resistance normally increases with decreasing current levels.

40. Measurement Techniques - Ammeters • For an up-scale (analog meter) or positive (digital meter) reading, an ammeter must be connected with current entering the positive terminal (Red) of the meter and leaving the negative terminal (Black), as shown in fig.

41. Measurement Techniques - Voltmeters • Voltmeters are always hooked up across the element for which the voltage is to be determined. • An up-scale or positive reading on a voltmeter is obtained by connecting positive terminal (red lead) to the point of higher potential and the negative terminal (black lead) is connected to the lower potential

42. Measurement Techniques - Voltmeters

43. Measurement Techniques - Voltmeters • The internal resistance of a supply cannot be measured with an ohmmeter due to the voltage present.

44. Measurement Techniques - Voltmeters • However, the no-load voltage can be measured by simply hooking up the voltmeter as shown. • The internal resistance of the voltmeter is usually sufficiently high to ensure that the resulting current is so small that it can be ignored.

45. Measurement Techniques • It seems we can find by Ohm’s law: Rint = ENL /ISC. • However, Rint of the voltage supply may be very low, resulting in high current levels which could damage the ammeter and supply. • A better approach would be to apply a RL resistive load and then measure current and use following eq. To calculate Rint.

46. Applications – Holiday Lights • If one wire enters and leaves the bulb casing, they are in series. • If two wires enter and leave, they are probably in parallel.

47. Applications – Holiday Lights • When bulbs are connected in series, if one burns out (the filament breaks and circuit opens), and all the bulbs should go out

48. Applications – Holiday Lights • However, the holiday bulbs are specially designed to permit current to continue to flow to the other bulbs when the filament burns out. • Note that only one flasher unit is required per 50 bulb panel

49. Applications – Holiday Lights • The bulbs of fig are rated 2.5 V at 0.2 A or 200 mA. • Since there are 50 bulbs in series, the total voltage will be 50 x 2.5 V = 125 V which matches voltage available • Since the bulbs are in series, the current through each bulb will be 200 mA. • The power rating of each bulb is therefore P = VI (2.5 V)(0.2 A) = 0.5 W with a total wattage demand of 50 x 0.5 W = 25 W.

50. Applications – Holiday Lights • When each set is connected together, they will actually be in parallel • Note that the top line is the hot line to all the connected sets, and the bottom line is the return, neutral, or ground line for all the sets.