EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
Lecture 3
• Definitions: Circuits, Nodes, Branches
• Kirchoff’s Voltage Law (KVL)
• Kirchoff’s Current Law (KCL)
• Examples and generalizations
• RC Circuit Solution
1
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
BRANCHES AND NODES
Branch:
elements connected end-to-end,
nothing coming off in between (in series)
Node:
place where elements are joined—entire wire
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
NOTATION: NODE VOLTAGES
The voltage drop from node X to a reference node
(ground) is called the node voltage Vx.
Example:
a
+
Va
_
b
+
+
_
Vb
_
ground
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KIRCHOFF’S VOLTAGE LAW (KVL)
The sum of the voltage drops around any closed loop is zero.
We must return to the same potential (conservation of energy).
Path
Path
+
V1 “drop”
-
- “rise” or “step up”
V2 (negative drop)
+
Closed loop: Path beginning and ending on the same node
Our trick: to sum voltage drops on elements, look at the first sign
you encounter on element when tracing path
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KVL EXAMPLE
+ v2 
1
+
va

b

a
+
vb
-
v3
2
3
Path 1:
 v a  v 2  vb  0
Path 2:
 vb  v3  vc  0
Path 3:
 va  v2  v3  vc  0
+
Examples of
three closed
paths:
c
+
vc

EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
UNDERLYING ASSUMPTIONS OF KVL
Assume no time-varying magnetic flux through the loop …
If there was, Faraday’s Law  induced emf (voltage)
Antennas are designed to “pick up”
electromagnetic waves
“Regular circuits” often do the same thing 
not desirable!

B( t )

+
v( t )
Avoid these loops!
How do we deal with antennas (EECS 117A)?
Include a voltage source as the circuit representation of the
emf or “noise” pickup.
We have a lumped model rather than a distributed (wave) model.
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
ALTERNATIVE STATEMENTS OF KIRCHHOFF’S
VOLTAGE LAW
1) For any node sequence A, B, C, D, …, M around a
closed path, the voltage drop from A to M is given by
v AM  v AB  v BC  v CD    v LM
2) For all pairs of nodes i and j, the voltage drop from i to j is
v ij  v i  v j
where the node voltages are measured with respect to
the common node.
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
MAJOR IMPLICATION
KVL tells us that any set of elements which are connected at
both ends carry the same voltage.
We say these elements are in parallel.
KVL clockwise,
start at top:
Vb – Va = 0
Va = Vb
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KIRCHOFF’S CURRENT LAW
Circuit with several branches connected at a node:
i2
i1
i3
i4
KIRCHOFF’s CURRENT LAW “KCL”:
(Sum of currents entering node)  (Sum of currents leaving node) = 0
Charge stored in node is zero (e.g. entire capacitor is part of a branch)
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
USING KCL
Kirchhoff’s Current Law (KCL)
Formulation 1:
Sum of currents entering node = sum of currents
leaving node
Use/write reference directions to determine
“entering” and “leaving” currents--no concern
about actual current directions
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
ALTERNATIVE KCL FORMULATIONS
Formulation 2:
“Algebraic sum” of currents entering node = 0
where “algebraic sum” means currents leaving are
included with a minus sign
Formulation 3:
“Algebraic sum” of currents leaving node = 0
currents entering are included with a minus sign
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
MAJOR IMPLICATION
KCL tells us that all of the elements in a single branch carry
the same current.
We say these elements are in series.
Current entering node = Current leaving node
i1 = i 2
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KIRCHHOFF’S CURRENT LAW EXAMPLE
24 A
-4 A
10 A
i
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
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
GENERALIZATION OF KCL
Sum of currents entering/leaving a closed surface 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.
The surface can enclose more than one node!
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KIRCHOFF’S CURRENT LAW USING SURFACES
Another example
Example
surface
50 mA
5 A  2 A  i
5A
2A
entering
i
i=7A
leaving
i?
i must be 50 mA
EECS 40 Spring 2003 Lecture 3
W. G. Oldham and S. Ross
KCL EXAMPLE: RC CIRCUIT
R
KCL at node X:
Current into X from the left: Vin
(Vin - Vx) / R
+
_
Current out of X down to ground:
X
+
C
Vx
ground
C dVx / dt
KCL: (Vin – Vx) / R = C dVx / dt

Solution:
dVout
1

( Vin  Vout )
dt
RC
Vx(t) = Vin + [ Vx(t=0) – Vin ] e-t/(RC)