Presentation is loading. Please wait.

Presentation is loading. Please wait.

ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 11 BY MOEEN GHIYAS.

Similar presentations


Presentation on theme: "ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 11 BY MOEEN GHIYAS."— Presentation transcript:

1 ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 11 BY MOEEN GHIYAS

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

3 Today’s Lesson Contents Kirchhoff’s Voltage Law (KVL) Interchanging Series Elements Voltage Divider Rule Notation

4 Kirchhoff’s Voltage Law Kirchhoff’s voltage law (KVL) states that the algebraic sum of the potential rises and drops around a closed loop (or path) is zero. A closed loop is any continuous path that leaves a point in one direction and returns to that same point from another direction without leaving the circuit.

5 Kirchhoff’s Voltage Law To apply Kirchhoff’s voltage law, the summation of potential rises and drops must be made in one direction (whether clockwise or anticlockwise) around the closed loop. In symbolic form, where Σ represents summation, the closed loop, and V the potential drops and rises, we have

6 Kirchhoff’s Voltage Law Reveals that, the applied voltage of a series circuit equals the sum of the voltage drops across the series elements For clockwise direction, For anticlockwise direction,

7 Kirchhoff’s Voltage Law The application of Kirchhoff’s voltage law need not follow a path that includes current-carrying elements. Using the clockwise direction

8 Kirchhoff’s Voltage Law Example – Determine the unknown voltages for the network of fig. Solution: For clockwise direction,

9 Kirchhoff’s Voltage Law Example – Determine the unknown voltages for the network of fig. Solution: Two possible loops; using clockwise direction, Loop 1 – including source E,

10 Kirchhoff’s Voltage Law Example – Determine the unknown voltages for the network of fig. Solution: Two possible loops; using clockwise direction, Loop 2 – including R 2 and R 3

11 Kirchhoff’s Voltage Law (KVL) Example – Using Kirchhoff’s voltage law, determine the unknown voltages for the network of fig. Solution:

12 Kirchhoff’s Voltage Law (KVL) Example – Using Kirchhoff’s voltage law, determine the unknown voltages for the network of fig. Note the polarity of the unknown voltage is not provided Solution: Since the result is negative, we know that a should be negative and b should be positive, but the magnitude of 18 V is correct.

13 Interchanging Series Elements The elements of a series circuit can be interchanged without affecting the total resistance, current, or power to each element. For instance, the network of figs. –R T = 35 Ω, and I =70V / 35Ω = 2 A in both cases –V 2 = IR 2 = (2 A)(5 ) = 10 V for both configurations

14 Interchanging Series Elements Example – Determine I and the voltage across the 7Ω resistor for the network of Fig.. Solution:

15 Voltage Divider Rule In a series circuit, the voltage across resistive elements will divide as the magnitude of the resistance levels. The largest resistor R 1 = 6Ω captures the bulk of the applied voltage, while the smallest resistor R 3 = 1Ω has the least.

16 Voltage Divider Rule In simple words, the voltage across series resistors will have the same ratio as their resistance levels. Note:Since resistance level of R 1 is 6 times that of R 3, the voltage across R 1 is 6 times that of R 3. Also R 2 / R 3 = 3, so does V 2 / V 3 = 3 Finally, since R 1 is twice R 2, the voltage across R 1 is twice that of R 2.

17 Voltage Divider Rule Note that, if the resistance levels of all the resistors of fig (left) are increased by the same amount, as shown in fig (right), the voltage levels will all remain the same.

18 Voltage Divider Rule Therefore, it is the ratio of resistor values that counts when it comes to voltage division and not the magnitude of the resistors.

19 Voltage Divider Rule The current level of the network will be severely affected by the change in resistance level from fig (left) to fig (right), but the voltage levels will remain the same.

20 Voltage Divider Rule - Example Note here: 1 MΩ = (1000)1 kΩ = (10,000)100Ω revealing that V 1 = 1000V 2 = 10,000V 3. Check:

21 Voltage Divider Rule In the last slide / discussion the current was determined before the voltages of the network were determined There is, however, a method referred to as the voltage divider rule (VDR) that permits determining the voltage levels without first finding the current

22 Voltage Divider Rule The rule can be derived by analyzing the network of fig. Applying Ohm’s Law,

23 Voltage Divider Rule The voltage divider rule states that the voltage across a resistor in a series circuit is equal to the value of that resistor times the total impressed voltage across the series elements divided by the total resistance of the series elements. where V x is the voltage across R x, E is the impressed voltage across the series elements, and R T is the total resistance of the series circuit.

24 Voltage Divider Rule Example – Determine the voltage V 1 for the network. Solution: We know or

25 Voltage Divider Rule Example – Determine the voltage V′ in fig across resistors R 1 and R 2. Solution: We know or

26 Voltage Divider Rule There is also no need for the voltage E in the equation to be the source voltage of the network. For example, if V is the total voltage across a number of series elements such as those shown in above fig, then

27 Voltage Divider Rule Example – Design the voltage divider of fig such that V R1 = 4V R2. Solution: The total resistance is Then.Therefore.We have.and

28 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.

29 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)

30 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)

31 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

32 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 V ab, point a must have a higher potential than point b if V ab = +ve value. If point b is at a higher potential than point a, then V ab = -ve value.

33 Notation – Single-Subscript Notation The single-subscript notation V a 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 V b If the voltage is less than zero volts, a negative sign must be associated with the magnitude of V a

34 Notation – General Comments Also the voltage V ab can be determined using Eq. V ab = V a – V b For fig below:

35 Notation Example – Find the voltage V ab 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.

36 Notation Example – Find voltage V a for the configuration of Fig Solution:

37 Notation Example – Find the voltage V ab for the configuration. Solution:

38 Notation & Voltage Divider Rule Example – Using the voltage divider rule, determine the voltages V 1 and V 2 of fig. Solution:Circuit Redrawn,. From.voltage divider rule,

39 Notation & Voltage Divider Rule Example – For the network of fig a)Calculate V ab. b)Determine V b. c)Calculate V c.

40 Notation & Voltage Divider Rule a)Calculate V ab. Solution: c)Determine V c. Solution:

41 Notation & Voltage Divider Rule b)Determine V b. Solution: or

42 Solution to Problems #1d – Find the total resistance and current I for given circuit Solution: R T = 3k Ω + 1.3k Ω + 4.5k Ω + 1.2k Ω R T = 10kΩ

43 Solution to Problems #8b – Determine the unknown voltages using Kirchhoff’s voltage law. Solution: Loop1: 24V – 10V – V 1 = 0 V 1 = 14V Loop2: 10V – V 2 + 6V = 0 V 2 = 16V

44 Solution to Problems #16b – Find the unknown resistance using the voltage divider rule and the information provided for the fig. Solution: Using VDR ratio method V 3Ω = 60V V 6Ω = 120V V R = 140 – 120V = 20V Since V 6Ω = 6 V R, therefore R 6Ω = 6 R.R = 6Ω/6 = 1Ω

45 Solution to Problems #32b – Determine the voltages V a, V b, and V ab for the network Solution: Va = = 4V Vb = -8V Vab = 12V

46 Summary / Conclusion Kirchhoff’s Voltage Law (KVL) Interchanging Series Elements Voltage Divider Rule Notation

47


Download ppt "ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 11 BY MOEEN GHIYAS."

Similar presentations


Ads by Google