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1
EE 215A Fundamentals of Electrical Engineering
Lecture Notes
Resistive Circuits
10/06/04
Rich Christie
Introduction:
The solution of circuits with more than two elements needs a little more theory. Start with
some definitions:
Nodes - points of connection of the circuit.
Wires - ideal wires have zero resistance, are short circuits.
Can expand a note to a wire.
1
1
2
+
2
+
-
3
3
Can collapse a wire to a node.
How many nodes?
+
-
ic
215A
2
+
-
+
+
-
-
Kirchhoff's Current Law (KCL):
A node is just a point - it can't store charge. So charge flowing in to a node must flow out
of it. Flow of charge is just current, of course. A clever way to say this is:
"The sum of currents flowing into a node is zero."
(We could also say the sum of currents flowing out of a node is zero. But we say "into"
because a guy named Kirchhoff said into, and got his name on the saying as)
Kirchhoff's Current Law (KCL).
!i
i
=0
i3
i1
i2
i1 + i 2 + i3 = 0
i3
i1
i2
i4
3
Suppose I add a current coming out? It's just the negative of the current flowing in
i1 + i 2 + i3 + (! i 4 ) = 0
Or you can think of the sum of current in equal to the sum of current out (more natural less general)
i1 + i 2 + i3 = i 4
i3
3A
-2A
-6A
What is i3?
3+(-2)-(-6)-i3 = 0
i3 = 7A
Or, since i3 flows out, it is the sum of what flows in, which is 3 -2 is 1 and -6 out is 6 in
plus 1 is 7 A out…
It’s good to develop the ability to do KCL around a node for any combination of
polarities and signs of current values.
Kirchhoff's Voltage Law (KVL):
Start with another definition:
Loops - paths that follow branches (circuit elements) from node to node until they
end up back at the starting point.
Count the loops:
+
-
ic
4
+
+
+
-
It should be obvious that
"The sum of the voltages around a loop is zero."
It was obvious to Kirchhoff, and he got his name on another law, Kirchhoff's Voltage
Law, KVL.
+
8V
Let's go around the loop clockwise from lower
left.
-
+3V - 8V + 5V = 0 (checks)
3V
+
-
-
+
5V
Note how polarity of voltage is important!
To deal correctly with polarity:
1. Decide which direction you will trace the loop.
2. Decide whether you will use the near side or the far side polarity as the sign for each
term.
IT DOES NOT MATTER WHICH CHOICE YOU MAKE AS LONG AS YOU APPLY
IT CONSISTENTLY.
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In the example above, we went clockwise and used the far sign. Let's go clockwise and
use the near sign. We get
-3V + 8V - 5V = 0 V (same answer).
For practice, try counterclockwise near sign and counterclockwise far sign.
Instead of deciding each time which direction and which sign, make a choice now. I
recommend clockwise and far sign, it seems the most natural to me. Then use those
choices unless there is a good reason not to.
+
4V
-
-5 V
+
3V
-
v
+
-
+
-
-
+
-2 V
What's the value of v?
Write KVL clockwise starting at lower left:
+3 - 4 - (-5) + v + (-2) = 0
v = -2 V
As with KCL, we can write the sum of the voltages around part of the loop equal to the
sum of the voltages around the other part. Thus, take a loop from the (-) terminal of v
above clockwise one element, and set it equal to the voltage around the rest of the loop
going counterclockwise from the same starting point to the same ending point:
+v = +(-5) + 4 - 3 - (-2) = -2V
With these laws plus Ohm's law we can solve (find all voltages and currents) in any
resistive circuit. We'll spend a lot of time learning different ways to do this.
Series Resistors:
We often find resistors in series. (Connected to the same node at one end, and different
nodes at opposite ends. Nothing can be connected to the middle node.) The same current
flows through both resistors (a basic test for series connection).
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i
+
v1
vs
There are two things that are easy to do when you
see a two (or more) resistors in series. They are:
+
R1
+
-
v2
R2
-
1. Simplify the circuit by replacing the two (or
more) resistors with one series equivalent
resistor.
2. Knowing the voltage across the series
combination, find the voltage across each of the
resistors.
Note that you can't do both at the same time! If
you combine resistors, the node between them
goes away!
Let's do replacement first.
We want to find Req such that current i is
unchanged, for a given voltage vs.
i
From KVL on the original circuit
vs
+
Req
-
v s ! v1 ! v 2 = 0
v s = v1 + v 2
v s = iR1 + iR2 (Ohm's Law)
v s = i (R1 + R2 ) = iReq
Req = R1 + R2
It should be pretty clear that when you see resistors in series, you can just add them right
up.
Resistors in series add!
What about finding the voltages across the resistors? From above
v s = i (R1 + R2 )
vs
R1 + R s
v1 = iR1
i=
but from Ohm's Law
so
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v1 = v s
R1
R1 + R2
similarly
v2 = vs
R2
R1 + R2
A voltage across series resistors divides among the resistors in proportion to the
resistance of each - hence the term Voltage Divider. If there are more than two resistors
in series, the sum of all the resistances appears in the divisor (the bottom of the fraction.)
The larger resistors will have more of the voltage across them.
Examples:
v2 is 6V, Req is 10 Ω
+
v1
10 V
+
4!
+
-
v2
6!
-
Find v2, power supplied by source
1 k!
9V
v 2 = !9
-
+
v2
-
+
1.5 k!
500
= !1.5V
1000 + 500 + 1500
Power:
500 !
Req = 3k!
v2
92
p=
=
= 27 mW
R 3000
8
OR
9
= 3 mA
3k
p = iv = 3 ! 9 = 27 mW
i=
Could also find v across each resistor and p dissipated by each resistor.
Could also find p dissipated in each resistor using current
Could also find p dissipated in Req using current
Parallel Resistors:
We often find resistors in parallel - connected to same node at both ends. Voltage across
resistors is the same.
is
Can we replace? Yes.
i1
vs
+
-
R1
i2
vs
vs
=
is
i1 + i 2
vs
=
vs vs
+
R1 R2
Req =
R2
Req
Req =
1
=
R1 R2
R1 + R2
1
1
+
R1 R2
Pragmatic: For two parallel resistors, know it is product over sum - or sum over product,
hmm, units tell you which (product over sum is correct).
For three or more resistors, use "one over one over plus"…
Req =
1
1
1
1
+
+
+ ...
R1 R2 R3
As you put in more parallel resistors, the equivalent resistance goes down.
It is always less that the smallest resistor in the parallel set.
Note: Parallel conductances add: G eq = G1 + G 2 + ...
Like voltage divider there is a current divider. The most common case is two resistors:
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v s = i s Req = i s
ii R1 = i s
i1 = i s
R1 R2
R1 + R2
R1 R2
R1 + R2
R2
R1 + R2
Note the resistor you are NOT interested in appears in the numerator (top). Think about
it, more current flows through the smaller resistor - so the bigger one appears on the top.
For more than two parallel resistors you have to use conductances.
in = i s
1
Rn
Gn
= is
1
1
1
G eq
+
+
+ ...
R1 R2 R3
Example
15
= 3.6 A
10 + 15
i 2 = 6 ! 3.6 = 2.4 A
10
i2 = 6
= 2.4 A check!
10 + 15
i1 = 6
i1
6A
10 !
i2
15 !
Sources:
Series voltage sources simply add (KVL)
Parallel current sources simply add (KCL)
Parallel voltage sources of different values are theoretically impossible. If you try this in
practice, you will hurt something (melt, explode or get very hot).
Series current sources of different values have same effect - impossible in theory,
melting, explosions or excess heat in practice. Ouch!
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Summary:
Tools you now have to attack ckts:
•
•
•
•
•
•
•
Ohm's Law
KCL
KVL
Series combination
parallel combination
voltage divider
current divider.
Challenge is to pull right tool out of toolkit, apply to right part of circuit, to get to answer
most easily (fastest, least likely to make an error - both depend on the skills of the solver
- you!). When doing this in an unstructured way, it is called
"Opportunistic circuit theory"
Bit of a logic problem - need to practice, practice, practice solving circuits. Eventually it
clicks. There is no step by step process for choosing the shortest correct sequence of
steps. (Can talk AI problem solver if you want.) There are, however, systematic methods
of solution that we'll cover in the next few lectures, nodal and mesh analysis.