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Chapter 10 The Math for Kirchhoff Voltage and Current Laws along with Polarity in DC Circuits.

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Presentation on theme: "Chapter 10 The Math for Kirchhoff Voltage and Current Laws along with Polarity in DC Circuits."— Presentation transcript:

1 Chapter 10 The Math for Kirchhoff Voltage and Current Laws along with Polarity in DC Circuits

2 Kirchhoff Laws Overview  There are three basic laws that cover Voltages, Currents, and Resistances in circuits. These laws have basic equations that describe them. This chapter covers Kirchhoff’s Voltage and Current Laws. Ohm’s Law is covered in Chapter 11. A topic closely related to solving the equations related to these laws, Circuit Polarity is also covered in this chapter. Kirchhoff Current Law  Current flow basics See Figure 10-3 on page 200 The number of charges (current) flowing to a point in a conductor must equal the number of charges flowing from the point

3 Kirchhoff Laws Kirchhoff Current Law  Current flow basics The algebraic sum of currents into and out of a point must equal zero if the currents towards a point are given an arbitrary sign of ‘+’ and currents leaving the point an arbitrary sign of ‘-’ See Fig 10-3.a on page 200)  Current in a series circuits There is only one path for free electrons to follow, thus it is the same any where in the circuit.

4 Kirchhoff Laws Kirchhoff Current Law  Current in a series circuits When a series circuit is connected across a voltage source, the free electrons must drift through all the series resistances If there are two or more resistances in the same current path, the total resistance across the voltage source is the sum of all the resistances

5 Kirchhoff Laws Kirchhoff Current Law  Current in a parallel circuits See Figure 10-5 on page 201 The sum of currents towards a junction equals the sum of currents leaving a junction –The algebraic sum of currents into a junction and out of a junction equals zero if the if the currents towards a junction are given an arbitrary sign of ‘+’ and currents leaving the junction an arbitrary sign of ‘-’ Example - See Figure 10-7 on page 202 –Currents into the junction (I 1 and I 2 ) sum to 5 amps –I 3 is the current leaving the junction and must equal 5 amps –I 1 + I 2 = I 3 –I 1 + I 2 - I 3 = 0

6 Kirchhoff Laws Kirchhoff Current Law  Current in a parallel circuits Example - See Figure 10-8 on page 202  Currents into the junction is I 1 and equals 8A. Currents out of the junction are I 2 (1A), I 3 (3A), I 4 and must equal 8A.  I 1 - I 2 - I 3 – I 4 = 0,  8A -1A -3A – I 4 = 0  8A -1A -3A = I 4  4A = I 4  Another I T = I 1 + I 2 + I 3 + I 4 = I T I1I1 V I2I2 I3I3 I4I4 ITIT ITIT

7 Kirchhoff Laws Kirchhoff Current Law  Current in a series-parallel circuits Overview –Circuits with both series and parallel characteristics is known as series-parallel circuits –See Figure 10-11 on page 203  I T comes out of the source splits into I 2 and I 3 at point B  At point C I 2 and I 3 recombine into I T  I T flows through Lamp 1 and back into the source –Note Current out of a source must equal current into that source Another Example - See Figure 10-12 on page 204 –Note current through components are given the same subscript as the components – See I 4 and R 4 – Application of Kirchhoff’s Current Law yields:

8 Kirchhoff Laws Kirchhoff Current Law  Current in a series-parallel circuits Another Example - See Figure 10-12 on page 204 –Application of Kirchhoff’s Current Law yields:  I T = I 1  I T = I3 + I 4  I T = I2 + I 4  I 2 = I 3  I 1 = I 2 + I 4 Use Example 10-1 on page 204

9 Kirchhoff Laws Kirchhoff Current Law  Current in a series-parallel circuits Use Example 10-1 on page 204 Do Example 10-2 on page 205

10 Kirchhoff Laws Kirchhoff Voltage Law  Overview If an electrical charge travels around any closed path and returns to its starting point, its net potential energy change must be equal to zero. The algebraic sum of emf’s (E) and voltage drops (V) around a closed path is equal to zero (ΣE – ΣV = 0)

11 Kirchhoff Laws Kirchhoff Voltage Law  Example See Figure 10-18 at the bottom of page 207 –The emf source (E) equals 10V and the voltage drops across Lamp 1 (L 1 ) and Lamp 2 (L 2 ) equals 6V and 4V –Thus E - V 1 - V 2 = 0, 10V – 6V -4V = 0 Do examples 10 – 3, 10 – 4, and 10 – 5 on pages 208 and 209  Example 10 – 6 Given: –E=10V –V 2 = 5V –V 3 = 2V Find: V 1 and V 4

12 Kirchhoff Laws Kirchhoff Voltage Law  Example 10 – 6  Do Example 10 – 7 on page 210

13 Polarity Overview  In electrical circuits, the polarity of the voltage ddrops across the individual resistances depends on the direction of current through them.  There are two different conventions for flow of current: Electron flow. Current flows from the negative side of the source. It enters the ‘-’ side of resistors and exits on the ‘+’ side of the resistor. Conventional Current flow. Current flows from the positive side of the source. It enters the ‘+’ side of the side of the resistors and exits on the ‘-’ side of the resistor.

14 Polarity Overview  Regardless of the convention used. It must be used consistently – conventions cannot be mixed The algebraic sum of emf’s and voltage drops must equal zero See figures 10-25, 10-26, and 10-27 Do Example problem 10-8 on page 213 Practice with the odd numbered problems at the end of the chapter.


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