Lecture 2 Kirchhoffs Laws
Definitions:
Branch: elements connected end-to-end, nothing coming off in between (in series)
Node:
place where elements are joinedentire wire
Loop: a closed path formed by tracing through an ordered sequence of nodes without passing through any node more than once. If we start at any point in a circuit (node), proceed through connected electric devices back to the point (node) from which we started, without crossing a node more than one time, we form a closed-path.
vb R2 + v2 va a + v1 R1 b + v3 R3
Kirchhoffs Voltage Law(KVL)
Kirchhoffs voltage law - how to handle voltages in an electric circuit. Kirchhoffs voltage law - the algebraic sum of the voltages around any closed path (loop) equal zero. 3 ways to interrupt KVL
KVL (cont.)
Consideration 1: Sum of the voltage drops around a circuit is equal to zero. We first define a drop. We assume a circuit of the following configuration. _ v2 + + v1 _ _ v3 + + _ v4
KVL (cont.)
We define a voltage drop as positive if we enter the positive terminal and leave the negative terminal. + v1 _ The drop moving from left to right above is + v1. _ v1 +
The drop moving from left to right above is v1.
KVL (cont.)
Consider the circuit loop of the following. If we sum the voltage drops along the loop, starting at point a, we write:
drops in CW direction starting at a - v1 v2 + v4 + v3 = 0 _ v2 +
+ v1 _
a
+ _ _ v3 +
drops in CCW direction starting at a
v4
- v3 v4 + v2 + v1 = 0
KVL (cont.)
Consideration 2: Sum of the voltage rises along a loop is equal to zero. We first define a rise. We define a voltage rise in the following diagrams: _ v1 +
The voltage rise in moving from left to right above is + V1. + v1 _ The voltage rise in moving from left to right above is - V1.
KVL (cont.)
Consider the circuit of the following. If we sum the voltage rises along the loop, starting at point a we write:
+ v1 + v2 - v4 v3 = 0
_
rises in the CW direction starting at a
v2 + + _ v4
+ v1 _
a
v3 +
rises in the CCW direction starting at a
+ v3 + v4 v2 v1 = 0
KVL (cont.)
For any given circuit, there are a fixed number of loops for writing KVL equations that are linearly independent.
Both the starting point and direction of the loop to write KVL equation are arbitrary. However, one must end the path at the same point from which one starts. In most text, the sum of the voltage drops equal to zero is normally used in applying KVL.
For the circuit of the following, there are a number of closed loops.
v1 +
+ v 2
v4 +
- v5 +
v6 + Path 1 Path 2 v8 + Path 3
v3
+ + v10 -
+ v7 -
+ v12 + v11 -
v9 +
b v1 +
Using sum of the V drops = 0 + v 2 v4 + v3 + + v10 + v11 v9 + + v7 - v5 + v6 + Blue path, starting at a
- v7 + v10 v9 + v8 = 0
a
v8 + Red path, starting at b
+ v12 -
+v2 v5 v6 v8 + v9 v11 v12 + v1 = 0
Green path, starting at b
+ v2 v5 v6 v7 + v10 v11 - v12 + v1 = 0
Kirchhoffs Voltage Law: Double Subscript Notation
Voltages in circuits are often described using double subscript notation. Consider the following:
Vab : means the potential of point a with respect to point b with
point a assumed to be at the highest (+) potential and point b at the lower (-) potential.
Kirchhoffs Voltage Law: Double Subscript Notation
Write Kirchhoffs voltage law going in the clockwise direction for the diagram in the following Figure:
b
Going in the clockwise direction, starting at b, using rises;
vab + vxa + vyx + vby = 0
Kirchhoffs Voltage Law: Equivalences in voltage notations
The following are equivalent in denoting polarity. a
vab = v1
v1
-
v1
=
b
Assumes the upper terminal is positive in all 3 cases
v2
v2 = - 9 volts means the right hand side of the element is actually positive.
Kirchhoffs Current Law(KCL) Law(KCL)
The sum of all currents entering a node is zero, or
i1(t) i2(t)
i5(t) i4(t)
i3(t)
The sum of currents entering node is equal to sum of currents leaving node.
i (t ) = 0
j =1 j
KCL Example
24 A 10 A -4 A
Currents entering the node: 24 A Currents leaving the node: 4 A + 10 A + i Three formulations of KCL:
1: 2: 3:
24 = 4 + 10 + i 24 ( 4 ) 10 i = 0 24 4 + 10 + i = 0
i = 18 A i = 18 A i = 18 A
GENERALIZATION OF KCL
Sum of currents entering/leaving a closed surface (super node) is zero
Could be a big chunk of circuit in here, e.g., could be a Black Box Note that circuit branches could be inside the surface (super node). The surface (super node) can enclose more than one node!
KIRCHOFFS CURRENT LAW USING SURFACES
Example
Surface (super node)
Another example
50 mA
5 A + 2 A = i
5A entering 2A i leaving i?
i=7A
i must be 50 mA
