Transcript Kirchhoffs Resistor Networks.pptx

Current Calculations
• Kirchhoff's Current Law
•
Kirchhoff's Current Law, also known as Kirchhoff's Junction Law and Kirchhoff's First
Law, defines the way that electrical current is distributed when it crosses through a
junction - a point where three or more conductors meet. Specifically, the law states
that:
•
The algebraic sum of current into any junction is zero.
•
The sum of current into a junction equals the sum of current out of the
junction.
• Since current is the flow of electrons through a conductor, it cannot
build up at a junction, meaning that current is conserved: what
comes in must come out. When performing calculations, current
flowing into and out of the junction typically have opposite signs.
This allows Kirchhoff's Current Law to be restated as:
• The sum of current into a junction equals the sum of current
out of the junction.
Kirchhoff's Current Law in
action
•
In the picture to the right, a junction of four conductors (i.e. wires) is shown. The
currents i2 and i3 are flowing into the junction, while i1 and i4 flow out of it. In this
example, Kirchhoff's Junction Rule yields the following equation:
i2 + i3 = i1 + i 4
Kirchhoff's Voltage Law
• Kirchhoff's Voltage Law describes the distribution of voltage within a
loop, or closed conducting path, of an electrical circuit. Specifically,
Kirchhoff's Voltage Law states that:
• The algebraic sum of the voltage (potential) differences in any
loop must equal zero.
• The voltage differences include those associated with
electromagnetic fields (emfs) and resistive elements, such as
resistors, power sources (i.e. batteries) or devices (i.e. lamps,
televisions, blenders, etc.) plugged into the circuit.
• Vb = V1 + V2
Resistor Networks
• Resistors in Series and Parallel
Identify which of these components are connected directly in series
with each other, and which are connected directly in parallel with each
other:
Answers
• Figure 1:
R2 in parallel with R3.
Figure 2:
R1 in series with R2.
Figure 3:
R2 in series with R3.
Figure 4:
R1 in series with R2; R3 in series with R4.
Figure 5:
R1 in parallel with R3; R2 in parallel with R4.
Figure 6:
R1 in series with R2.
Series/Parallel
Calculate the resistance between points A and B (RAB) for the following
resistor networks:
Answers
•
Figure 1:
RAB = 500 Ω
Figure 2:
RAB = 750 Ω
Figure 3:
RAB = 1.511 kΩ
Figure 4:
RAB = 940 Ω
Figure 5:
RAB = 880 Ω
Figure 6:
RAB = 80.54 Ω
Notes:
Note that the circuit in figure 4 is a "trick:" two of the resistors contribute absolutely nothing to R AB!
Be sure to discuss why this is with your students.
Discuss with your students how they approached each of these problems, and let the entire class
participate in the reasoning process. The point of this question, like most of the questions in the
Socratic Electronics project, is not merely to obtain the correct answers, but to stimulate
understanding of how to solve problems such as these.
Complete the table of values for this circuit:
Answers