The Laws of Electrical Networks
Aims:
• Introduce Kirchhoff’s laws.
• How to use them / tricks of the trade.
• Series and parallel circuits.
• Voltage and current division
Lecture 4
1
Networks
Problem: Determine the currents and voltages at all parts of a complex network of
resistors and batteries
This problem became important with the growth of electric telegraph and power
networks in the 19th C.
Now central to computer based
circuit simulators, SPICE etc.
Mesh
+
–
Node
Loop
+
A mesh is a loop
with no internal
branches
Mesh
–
Branch
Lecture 4
2
1
Kirchhoff’s Laws
(or rules)
Kirchhoff’s Current Law (KCL)
“The algebraic sum of all currents
entering a node, or closed surface, at
any instant of time is zero.”
Kirchhoff’s Voltage Law (KVL)
“The algebraic sum of the voltages
around a closed loop at any instant of
time is zero.”
[Hint: The problem is to keep track of
the signs!]
Lecture 4
Gustav Kirchhoff (1824 – 1887).
Also worked in Prussia. As well as
developing a treatment of electrical
networks he made fundamental
discoveries in optical radiation.
3
KCL (Current Law)
• Charge is a fundamental quantity and is conserved
• Charge cannot be created or destroyed
• Charge cannot accumulate at a node
• Strong electrostatic forces would stop current flowing
So in any time dt,
I=
dq
dt
I1
dt ( I1 + I 2 + I 3 + K + I n ) = 0
I2
In
or
∑I
n
=0
Remember the SIGN!
e.g. Currents entering are POSITIVE
Currents leaving are NEGATIVE
Lecture 4
I3
I4
I5
4
2
KVL (Voltage Law)
Total energy gained by a charge going round
a loop is
Sum of energy gain due to batteries – energy
loss in resistors
This must be ZERO or electrons would
continue to accelerate (or stop)!!
∑V = 0
R1 I
x
+V2
R2
+V1
Mesh current
Rn
R3
You must take great care with SIGNS!
Lecture 4
6
Lecture 4
7
Gravity again…
“KVL” for big dippers says that
however much you go up and
down in a complete loop you
always get off with the same
gravitational potential energy
(height) as you had at the start.
3
KVL (Voltage Law)
R1 I
Steps for doing a KVL calculation
1. Draw the loop. Mark the + and – terminals
of all batteries. Choose a current direction
(clockwise or ACW) and a starting point (X)
2. Starting from X in the direction of the
current:
• If you meet the a battery from the + side
ADD the voltage to the sum
• If you meet the a battery from the - side
SUBTRACT the voltage from the sum
• If you meet a resistor ADD IR to the
sum
• When you get back to X, stop and set the
sum to ZERO
x
+V2
R2
+V1
Mesh current
Rn
R3
−V1 + IR1 − V2 + IR2 + IR3 K + IRn = 0
If I comes out –ve, this simply
means that you guessed the
current direction wrong
Lecture 4
8
Resistors in Series
R1 I
V
R2
Rn
R3
⇓
I
V
V = IREQ
REQ
Lecture 4
9
4
Resistors in Series
R1 I
V
R2
Rn
R3
V = I ∑ Rn
So
⇓
I
V
(KVL)
V = IREQ
REQ
REQ = ∑ Rn
Resistors in series are can be replaced by a
single resistor with a value equal to their sum
REQ = R1+R2+R3+ ...+Rn
Lecture 4
10
Voltage Divider
I
Used to reduce a voltage:
VB
R1
V
R2
In general for n resistors in series the voltage across the nth resistor is :
Vn = VB ×
Rn
∑ Rn
Voltage divider rule
Lecture 4
11
5
Voltage divider
I
Used to reduce a voltage:
V = IR2
I=
VB
VB
R1 + R2
V = VB ×
R1
V
R2
R1 + R2
R2
In general for N resistors in series the
voltage across the nth resistor is :
Vn = VB ×
Rn
∑ Rn
Voltage divider rule
Lecture 4
12
Voltage dividers in practice:
VIN
“POTENTIOMETER”
VOUT
Volume controls, Brightness controls, joystick sensors
etc. etc.
• Volume controls use a “LOGARITHMIC TRACK”
because of the response of the ear
Resistance
R
EA
LIN
LOG
Rotation
Lecture 4
13
6
Resistors in Parallel
I
V
I
I1
I2
I3
In
R1
R2
R3
Rn
⇒
V
REQ
Resistors in parallel can be replaced by a singe resistor with a value equal to
the reciprocal of the sum of the reciprocals of individual resistors.
For two resistors
in parallel:
Lecture 4
14
Resistors in parallel
I
I
V
I1
I2
I3
In
R1
R2
R3
Rn
I = I1 + I 2 + I 3 + K + I n
I=
⇒
V
REQ
(KCL)
V V V
V
V
+
+ +K +
=
(Ohm's Law)
R1 R2 R3
Rn REQ
REQ =
1
1
1
1
1
+
+ +K +
R1 R2 R3
Rn
Resistors connected in parallel can be
replaced by a singe resistor with a
value equal to the reciprocal of the
sum of the reciprocals of individual
resistors
Lecture 4
15
7
Conductance
The quantity 1/R is called CONDUCTANCE
[Symbol in equations G or g, units Ω-1 or Siemens (S)]
Thus for resistors in parallel,
GEQ = G1 + G2 + G3 + K + GN
You can calculate the CONDUCTANCE of a resistor by
inverting the formula for resistance:
R=
ρl
A
∴ G=
Here σ is the conductivity of the
material, the reciprocal of
resistivity. σ = 1/ρ
Units are S m-1.
1 σA
=
R
l
Lecture 4
16
Current Divider
I
V
I
I1
I2
I3
In
R1
R2
R3
Rn
⇒
V
REQ
Resistors in parallel can be replaced by a singe resistor with a value equal to
the reciprocal of the sum of the reciprocals of individual resistors.
Lecture 4
17
8
Current Divider
I
V
I=
I
I1
I2
I3
In
R1
R2
R3
Rn
V
REQ
In =
V
Rn
So that
⇒
V
REQ
I n REQ
G
=
= n
I
Rn
GEQ
Resistors in parallel can be replaced by a singe resistor with a value equal to
the reciprocal of the sum of the reciprocals of individual resistors.
For two resistors
in parallel:
Lecture 4
18
Complex resistor networks
Networks of resistors can be simplified by repeated application of the series
and parallel rules.
•
•
•
Use the series rule for two resistors with no other
components joining at their common node
Use the parallel rule for resistors with both terminals
connected
Repeat until no further simplification is possible
Some networks cannot be simplified and require more powerful tools
Lecture 4
19
9