1 / 47

ELECTRICITY & MAGNETISM (Fall 2011)

ELECTRICITY & MAGNETISM (Fall 2011). LECTURE # 11 BY MOEEN GHIYAS. (Series Circuit – Chapter 5 ) Introductory Circuit Analysis by Boylested (10 th Edition). TODAY’S lesson. Today’s Lesson Contents. Kirchhoff’s Voltage Law (KVL) Interchanging Series Elements Voltage Divider Rule

nantai
Download Presentation

ELECTRICITY & MAGNETISM (Fall 2011)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

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

  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 R2 and R3

  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 bshould 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. • RT = 35 Ω, and I =70V / 35Ω = 2 A in both cases • V2 = IR2 = (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 R1 = 6Ω captures the bulk of the applied voltage, while the smallest resistor R3 = 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 R1 is 6 times that of R3, the voltage across R1 is 6 times that of R3. • Also R2 / R3 = 3, so does V2 / V3 = 3 • Finally, since R1 is twice R2, the voltage across R1 is twice that of R2.

  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 • V1 = 1000V2 = 10,000V3. • 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 thatthe 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 Vx is the voltage across Rx, E is the impressed voltage across the series elements, and RT is the total resistance of the series circuit.

  24. Voltage Divider Rule • Example – Determine the voltage V1 for the network. • Solution: • We know • or

  25. Voltage Divider Rule • Example – Determine the voltage V′ in fig across resistors R1 and R2. • 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 VR1 = 4VR2. • 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 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.

  33. 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

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

  35. 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.

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

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

  38. 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,

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

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

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

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

  43. Solution to Problems • #8b – Determine the unknown voltages using Kirchhoff’s voltage law. • Solution: Loop1: 24V – 10V – V1 = 0 V1 = 14V Loop2: 10V – V2 + 6V = 0 V2 = 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 V3Ω= 60V V6Ω= 120V VR = 140 – 120V = 20V Since V6Ω = 6 VR , therefore R6Ω = 6 R . R = 6Ω/6 = 1Ω

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

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

More Related