Transcript Document

Outline
• Resistances in Series and Parallel
• Network Analysis by Using Series and Parallel Equivalents
• Voltage-divider and Current-Divider Circuits
• Node-Voltage Analysis
• Mesh-Current Analysis
•Thevenin and Norton Equivalent Circuits
In a parallel ckt, the voltage across each element is the same.
Applying Ohm’s law, we can write
v
v
v
i1  , i2 
, i3 
R1
R2
R3
v
v
v
i  i1  i2  i3  

R1 R2 R3
1
1
1
i( 
 )v
R1 R2 R3
For your own practice!
Voltage-divider and current-divider ckts
Voltage Division
When a voltage is applied to a series
combination of resistances, a fraction
of a voltage appears across each of
the resistances.
vtotal
vtotal
i

Req R1  R2  R3
R1
v1  R1i 
vtotal
R1  R2  R3
R2
v2  R2i 
vtotal
R1  R2  R3
R3
v3  R3i 
vtotal
R1  R2  R3
Current Division
The total current flowing into
a parallel combination of
resistances divides, and a
fraction of the total current
flows through each resistance.
v  Reqitotal
R1 R2

itotal
R1  R2
v
R2
i1 

itotal
R1 R1  R2
v
R1
i2 

itotal
R2 R1  R2
Node-voltage analysis
Although they are very important concepts, series/parallel
equivalents and the current/voltage division principles are not
sufficient to solve all ckt problems. That’s how the node-voltage
analysis comes in.
A node is a point at which two or more ckt elements are joined
together. In node-voltage analysis, we first select one of the nodes
as the reference node. In principle, any node can be picked to be the
reference node.
Next, we label the voltages at each of the other nodes.
Write equations to solve for the voltages, then the current.
Writing KCL equations in terms of the node voltage:
to find the current flowing out of node n through a resistance
toward node k, we subtract the voltage at node k from the voltage
at node n and divide the difference by the resistance. For example,
If vn and vk are the node voltages and R is the resistance connected
between the nodes, the current flowing from node n toward node k
is given by
vn  vk
R
Note: the positive reference is at the head of the arrow.
Apply KCL at node 2, we get
v2  v1 v2 v2  v3


0
R2
R4
R3
Apply KCL at node 3, we get
v3  v1 v3 v3  v2


0
R1
R5
R3
Thevenin and Norton Equivalent ckts
In this section, we learn how to replace two-terminal ckts
Containing resistances and sources by simple equivalent ckts.
Thevenin Equivalent circuits
One type of equivalent ckts is the Thevenin equivalent, which
Consists of an independent voltage source in series with a resistance.
Thevenizing Procedure
1. Calculate the open-ckt voltage (Vth)
across the network terminals.
2. Redraw the network with each
independent source replaced by its
internal resistance. This is called “deactivation of the sources”.
3. Calculate the resistance (RTH ) of the redrawn network as seen
from the output terminals.
Different methods of finding RTH
(a) For independent sources: Deactivate the sources, i.e. for
independent current source, deactivate it by open circuiting its
terminals and for voltage source, deactivate it by shorting it. To
In case of non-ideal sources, the internal resistance will remain
Connected across the deactivated source terminals.
(b) For dependent sources in addition or in absence of independent
source.
(i) Find open circuit voltage Voc across the open circuited load
terminals. Next short circuit the load terminals and find the short
ckt current (Isc) through the shorted terminals.
The Thevenin’s equivalent resistance is then obtained as
RTh = Voc/Isc
Norton’s Theorem
According to this theorem, any two-terminal active network,
containing voltage sources and resistance when viewed from its
output terminals is equivalent to a constant current source and
an internal (parallel) resistance.
The constant current source (known as Norton’s equivalent current
source) is of the magnitude of the short circuit current at the
terminals.
Nortonizing Procedure
1. Remove RL, and short circuit the terminals a and b. The
current through the short circuited path is Isc or In.
2. Finding internal resistance of the network by deactivating
all the sources and replace them by its internal resistance. Then
Calculate RN of the redrawn network as seen from the output
Terminals.
3. Draw the Norton’s equivalent ckt.
Source Transformation
A voltage source in series with a resistance is externally
Equivalent to a current source in parallel with the resistance,
Provided that In = Vt/Rt
Maximum Power Transfer
Suppose that we have a two-terminal ckt and we want to connect a
Load resistance RL such that the maximum possible power is
Delivered to the load.
To analyze this problem, we replace the original ckt by its Thevenin
Equivalent as shown
vt
iL 
Rt  RL
pL  iL2 RL
Vt 2 RL
pL 
( Rt  RL ) 2
dpL Vt 2 ( Rt  RL ) 2  2Vt 2 RL ( Rt  RL )

0
4
dRL
( Rt  RL )
RL  Rt
Thus, the load resistance that absorbs the maximum power from a
Two-terminal ckt is equal to the Thevenin resistance.
The maximum power is given by
2
Vt
PL max 
4 Rt