Download - Chapter 10 The Math for Kirchhoff Voltage and Current Laws along with Polarity in DC Circuits
Chapter 10
The Math for Kirchhoff Voltage and Current Laws along with Polarity in DC Circuits
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
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.
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
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 (I1 and I2) sum to 5 amps
– I3 is the current leaving the junction and must equal 5 amps
– I1 + I2 = I3
– I1 + I2 - I3 = 0
Kirchhoff Laws• Kirchhoff Current Law
Current in a parallel circuits• Example - See Figure 10-8 on page 202
Currents into the junction is I1 and equals 8A. Currents out of the junction are I2 (1A), I3(3A), I4 and must equal 8A.
I1 - I2 - I3 – I4 = 0, 8A -1A -3A – I4 = 08A -1A -3A = I4 4A = I4
Another IT = I1 + I2 + I3 + I4 = IT
I1V
I2 I3 I4
IT
IT
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 203IT comes out of the source splits into I2 and I3 at point B
At point C I2 and I3 recombine into IT
IT 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 I4 and R4
– Application of Kirchhoff’s Current Law yields:
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: IT = I1
IT = I3 + I4
IT = I2 + I4
I2 = I3
I1 = I2 + I4
• Use Example 10-1 on page 204
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
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)
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 (L1) and Lamp 2 (L2) equals 6V and 4V
– Thus E - V1 - V2 = 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– V2 = 5V
– V3= 2V
• Find: V1 and V4
Kirchhoff Laws• Kirchhoff Voltage Law
Example 10 – 6
Do Example 10 – 7 on page 210
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.
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.