Electrical Technology

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 Nodes, Branches and Loops
 Circuit elements can be interconnected in
multiple ways.
 To understand this, we need to be familiar
with some network topology concepts.
 A branch represents a single element such
as a voltage source or a resistor.
 A node is the point of connection between
two or more branches.
 a path is a sequence of nodes
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 Nodes, Branches and Loops
 A loop is any closed path in a circuit.
 A loop is independent if it contains at least
one branch not shared by any other indep
endent loops.
 Two or more elements are in series if they
exclusively share a single node and thus ca
rry the same current.
 Two or more elements are in parallel if the
y are connected to the same two nodes an
d thus have the same voltage.
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 Nodes, Branches and Loops
these two networks are equivalent
there are three nodes and five branches
 Kirchhoff’s Laws
 Kirchhoff’s laws are useful in circuit analy
sis.
 There are two laws:
 Kirchhoff’s Current law (KCL)
 Kirchhoff’s Voltage law (KVL)
 Kirchhoff’s Current law (KC
L)
 Kirchhoff’s current law is based on cons
ervation of charge.
 It states that the algebraic sum of current
s entering a node (or a closed boundary) i
s zero.
 It can be expressed as:
 Kirchhoff’s Current law (KCL)
 KCL: Algebraic sum of currents entering
any node is zero.
 KCL: Alternative Forms
 Current IN is zero:

 Current OUT is zero:

 Current IN = OUT:
 Example of KCL Application
 Find the current through resistor R3 if it i
s known that the voltage source supplies
a current of 3 A.

Answer: i = 6 A
 Kirchhoff’s voltage law (KVL)
 Kirchhoff’s voltage law is based on cons
ervation of energy.
 It states that the algebraic sum of voltage
s around a closed path (or loop) is zero.
 It can be expressed as:
 Kirchhoff’s Voltage Law (KV
L)
 KVL: Algebraic sum of voltages around a
ny closed path is zero.
 KVL: Alternative Forms
 Sum of RISES is zero (clockwise from B):

 Sum of DROPS is zero (clockwise from

B):

 Two paths, sam

e voltage (A to
  B):
𝑣 =𝑣 + −𝑣
1 2 ( 3 )
 Example: Applying KVL
 Find vR2 (the voltage across R2) and the
voltage vx.

Answer: vR2 = 32 V and vx = 6 V.

 Applying KCL, KVL, Ohm’s L
aw
 Example: find the current ix and the volta
ge vx

Answer: vx = 12 V and ix = 120 mA

 Applying KCL, KVL, Ohm’s L
aw
 Example: solve for the voltage vx and an
d the current ix

Answer: vx = 8 V and ix = 1 A
 Series Connections
 Two or more elements are in series if the
y exclusively share a single node and thu
s carry the same current.
 Parallel Connections
 Two or more elements are in parallel if t
hey are connected to the same two node
s and thus have the same voltage.
 Example: Single Loop Circuit
 Calculate the power absorbed by each ci
rcuit element.

Answer:
p120V = −960 W, p30 = 1920 W
pdep = −1920 W, p15 = 960 W
 Example: Single-Node-Pair
 Find the voltage v and the currents i1 and i
2.

Answer: v = 2 V, i1 = 60 A, and i2 = 30 A
 Example: Single-Node-Pair
 Determine the value of v and the power supplie
d by the independent current source.

Answer: v = 14.4 V, power from current sour

ce is 345.6 mW
 Series and Parallel Sources
 Voltage sources c
onnected in serie
s can be combine
d into an equivale
nt voltage source:
 Series and Parallel Sources
 Current sources connected in parallel can
be combined into an equivalent current s
ource:
 Impossible Circuits

Vs in parallel (a) and Is in series (c) can lead

to “impossible circuits”
 Series Resistors
 If two or more resistors are
in series, then the same cu
rrent pass through them.
 Consider the circuit show
n:
 Applying Ohm’s law to b
oth resistors.

 If we apply K V L to the loop we hav

e:
 Series Resistors
 Combining the two equations:

 From this we can see there is an equivale

nt resistance of the two resistors:

 For N resistors in series:

 Series Resistors

Using KVL shows:

 Example: Circuit Simplifying
Find i and the power supplied by the 80 V source.

Answer: i = 3 A and p = 240 W supplied

 Parallel Resistors
• When resistors are in parallel, the voltage
drop across them is the same.

• By KCL, the current at node a is

• The equivalent resistance is:

 Resistors in Parallel

Using KCL shows:

 Two Resistors in Parallel

Req=𝑅1∥𝑅2
  Two resistors in parallel can be combi
ned using the

product / sum
shortcut.
Connecting resistors in parallel make
s the result smaller :
0.5 min(R1, R2) < R1 ‖ R2 < min(R1,R2)
 Voltage Division
• The voltage drop across any one resistor c
an be known.
• The current through all the resistors is the
same, so using Ohm’s law:

• This is the principle of voltage division.

 Voltage Division
Resistors in series “share” the applied voltag
e.
 Example: Voltage Division
Find vx

Answer: vx(t) = 4 sin t V

 Current Division
• Given the current entering the node, the voltage
drop across the equivalent resistance will be the
same as that for the individual resistors.

• This can be used in combination with Ohm’s la

w to get the current through each resistor:
 Current Division
Resistors in parallel “share” current through t
hem.
 Example: Current Division
Find i3(t)

Answer: i3(t) = 1.333 sin t V

 Parallel Capacitors
 We learned with resistors tha
t applying the equivalent ser
ies and parallel combination
s can simply many circuits.
 Starting with N parallel capa
citors, one can note that the
voltages on all the caps are t
he same.
 Applying KCL:
 Parallel Capacitors
 Taking into consideration the current voltage rel
ationship of each capacitor:

 Where
 From this we find that parallel capacitors combi
ne as the sum of all capacitance.
 From this we see that the parallel combination
of capacitors resembles the series combination
of resistors.
 Series Capacitors
 Turning our attention to
a series arrangement of
capacitors:
 Here each capacitor sha
res the same current.
 Applying KVL to the loo
p:

 Now apply the voltage

current relationship.
 Series Capacitors

 Where
 From this we see that the series combinat
ion of capacitors resembles the parallel c
ombination of resistors.
 Series and Parallel Capacitors
 Another way to think about the combinations of
capacitors is as follows:
 Combining capacitors in parallel is equivalent to
increasing the surface area of the capacitors:
 This would lead to an increased overall capacita
nce (as is observed).
 A series combination can be seen as increasing t
he total plate separation.
 This would result in a decrease in capacitance (a
s is observed).
 Series Inductors
 We now need to exte
nd the series parallel
combinations to ind
uctors.
 First, let’s consider
a series combination
of inductors.
 Applying KVL to the l
oop:
 Series Inductors
 Factoring in the voltage current relationship.

 Where

 Here we can see that the inductors have t

he same behavior as resistors.
 Parallel Inductors
 Now consider a parallel
combination of inducto
rs:
 Applying KCL to the circ
uit:

 When the current volta

ge relationship is consi
dered, we have:
 Parallel Inductors
• The equivalent inductance is thus:

• Once again, the parallel combination resembles

that of resistors.
 Wye-Delta Transformations
 There are cases wh
ere resistors are ne
ither parallel nor s
eries.
 Consider the bridg
e circuit shown her
e.
 This circuit can be
simplified to a thre
e-terminal equival
ent.
 Wye-Delta Transformations
 Two topologies ca
n be interchanged:
 Wye (Y) or tee (T) net
works.

Delta

networks.
 Transforming between t
hese two topologies oft
en makes the solution o
f a circuit easier.
 Wye-Delta Transformations
 The superimposed
wye and delta circui
ts shown here will u
sed for reference.
 The delta consists o
f the outer resistors,
labeled a,b, and c.
 The wye network ar
e the inside resistor
s, labeled 1,2, and 3.
 Delta to Wye
 The conversion formula for a delta to wye
transformation are:
 Wye to Delta
 The conversion formula for a wye to delta
transformation are: