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Transcript
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
Independent voltage source
Ideal source:
§ Across its terminals maintains voltage with prescribed waveform
§ It is able to deliver any current, including infinite, depending on load, to
preserve given voltage
§ It is able to deliver infinite power
Symbol
Example of the voltage time
function
Loading characteristics at
time instant tk
Actual voltage source:
§ Deliverable power is finite
§ Maximum value of the delivered current is limited
Symbol:
Load characteristic:
u(t) = ui(t) ¡ f [i(t)]
u(t) = ui (t) ¡ Ri i(t)
ui
ik
open-circuited
(internal) voltage
short-circuit
current
Loading characteristic of linear (thick black line) and example of
non-linear (thin blue curve) voltage source
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-1-
Power, delivered to the circuit by voltage source: Pu = U ¢ Iu
where Iu is current passing the voltage source (positive sign has current
outgoing positive terminal of the voltage source)
º can be negative (consuming power, e.g. accumulator in charger)
Independent current source
Ideal source:
§ Across its terminals maintains current with prescribed waveform
§ It is able to reach infinite voltage, depending on load, to keep delivering
given current
§ It is able to deliver infinite power
Loading characteristics at
time instant tk
Symbol:
Example of the current time
function
Actual current source:
§ Deliverable power is finite
§ Maximum value of the terminal voltage is limited
Symbol:
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
Load characteristic:
i(t) = ii(t) ¡ g[u(t)]
i(t) = ii (t) ¡ Giu(t)
-2-
ui
ik
open-circuited
(internal) voltage
short-circuit current
Loading characteristic of linear (thick black line) and example of nonlinear (thin blue curve) current source
!!! Warning !!! – These two connections from the point of view of output
terminals acts like an ideal sources
Ri
Ii
Ui
Ri
Power, delivered to the circuit by the current source: Pi = Ui ¢ I
where Ui is voltage across current source (positive orientation of voltage
is from the terminal, from which the current is flowing out)
º can be negative (consuming power)
Source compatibility
If both loading characteristics are the same, we cannot distinguish from the
point of view of output terminals (by measurement), if the source is actual
voltage source, or current source Æactual sources can be replaced by
compatible sources
150W
i
i
u
75V
u
up = ui ;
ik =
any
loading
circuit
ui
Ri
0.5A
150W
up =
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
ii
= iiRi;
Gi
any
loading
circuit
ik = ii
-3-
It is not possible to substitute following circuits – sources are still ideal!
Example:
I = 1A, R1 = 100 W,
R2 = 200 W, R3 = 300 W
Ui = IR2 = 200 V
Ri = R2 + R3 = 500 Ð
Thevenin’s theorem
§ Any linear active two-terminal („black box“, containing any number of
circuit elements – sources, resistors, inductors, capacitors, …) can be
replaced by the series connection of a voltage source and a passive
two-terminal (resistor, or connection of resistor, capacitor or inductor,
resulting on impedance)
§ The value of a passive two-terminal resistance (impedance) is total
resistance (impedance) of whole active two-terminal after removing
of all sources, from the point of view of terminals
§ Removing of voltage source: source is short-circuited (ideal voltage
source has zero internal resistivity)
§ Removing of current source: source is opened (ideal current source
has infinite internal resistivity /and not loaded has infinite voltage
across its terminals/)
§ It is not possible to remove controlled source from the circuit!
§ Total resistivity (impedance) is tangent of loading characteristics,
event the circuit contains controlled sources Æ the only method, if
the circuit contains controlled sources
Up
Ri =
Ik
§ Not valid for evaluation of total power of the circuit (different currents
and voltages) – power is not linear function!
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-4-
Norton’s theorem
§ Any linear active two-terminal can be replaced by the parallel
connection of a current source and a passive two-terminal
§ The value of a passive two-terminal resistance (impedance) is total
resistance (impedance) of whole active two-terminal after removing
of all sources, from the point of view of terminals
Controlled sources
§ the value of voltage (current) is controlled by another circuit variable
(voltage or current)
§ practical examples of devices containing controlled sources –
transistor, operational amplifier
voltage controlled voltage source current controlled voltage source
uv = Kur
uv = Rir
voltage controlled current source current controlled current source
iv = Hir
iv = Gur
Equivalence of active two-terminals
Series connection of voltage sources
n
X
u(t) =
uk (t)
k=1
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-5-
Parallel connection of current sources
n
X
i(t) =
ik (t)
k=1
≈
≈
Voltage source has 0 internal
resistivity, it is short circuit with
respect to current source and current
supplied by current source passes
voltage source freely; current source
has ∞ resistivity with respect to
voltage source, thus voltage source is
disconnected
Voltage source has 0 internal
resistivity, it is short circuit with
respect to current source and current
supplied by current source passes
voltage source; no current could be
delivered to the connected circuit;
current source has ∞ resistivity, it has
no effect on voltage source
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-6-
Voltage source replacement:
Current source replacement:
The left terminal of circuit elements D1,
D2 has voltage u(t); in the branch A the
voltage source is disturbed by another
voltage source connected in opposite
way; to maintan voltage u(t) in
branches B, C it is necessary connect
another two voltage sources u(t)
By connecting of second current
source in series it is possible to add
shorting strap to the C terminal;
current passing strap is 0 (whole
current pass only current sources)
so the conditions are same like
without strap
Dividing of the circuit:
If the circuit has two parts, connected together by two (or more) wires (these
are two or more terminals), it is possible to divide the circuit on two
independent parts, if the voltage conditions keep unchanged (we will connect
two voltage sources maintaining the same value of voltage)
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-7-
Example:
R1
R3
U1
R2
R4
→
R1
+
U1
R2
R3
U1
R4
Fundamental circuits
– voltage divider
· 2 or more circuit elements connected in series
· common circuit variable – current
U1
I=
R1 + R2
U1
U2 = R2I = R2
R1 + R2
R1
I
U1
R2
U2
U 2 = U1
Easy extendable to N resistors:
R2
R1 + R2
Uj = U1
– current divider
Rj
N
X
Ri
i=1
· 2 or more circuit elements connected in parallel
· common circuit variable – voltage
U = RI =
I
R1
I2 =
R2
U
R1 R2
I
R 1 + R2
U
R1 R2 1
=I
R2
R1 + R2 R2
I2 = I
R1
R1 + R2
Extension to N resistors is more complicated – e.g. for 3 resistors
R1 R2 R3
R = ( R11 + R12 + R13 )¡1 = R1 R2 +R
1 R 3 +R2 R3
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-8-
Step by step simplification method
· the aim is to simplify circuit by searching of series and parallel
connections of circuit elements starting from the output variable
· this procedure is repeated step by step until reaching some fundamental
circuit where it is easy to find voltage and passing current
· then, we will return back to the original circuit, step by step, dividing
voltages and currents
example:
Analyzed circuit
· The aim is compute voltage U2
R1
U1
R2
R3
R4
Step 1
We can combine series connected resistors R3
and R4 together
R1
U1
R2
R34
R1
U1
R234
U234
R1
U1
R3
R2
U234
U2
R4
U2
R34 = R3 + R4
Step 2
Now we will combine parallel connected resistors
R1 and R34;
Finally, resistors R1 a R234 form voltage divider, so
that we can find voltage across resistor R234
R2 ¢ R34
R234
R234 =
; U234 = U1
R2 + R34
R1 + R234
Step 3 – back to the original circuit
Now I know total voltage across resistors R3 a R4,
so that I can use voltage divider rule to find
voltage across resistor R4 – requested voltage U2
R4
U2 = U234
R3 + R4
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
-9-
Superposition theorem
R1
I
Ux
R1
R3
R2
U
I
Ux1
R3
R1
R2
Ux2
=
R3
R2
U
+
§ If the circuit is supplied by N independent voltage or current sources,
it is possible to remove N -1 sources, so that the circuit will be
supplied just by one source
§ Repeat previous step for each source in the circuit, so we obtain N
distinct results
§ Wanted circuit variable is the sum of N contributions from distinct
sources
§ NOT APPLICABLE if the circuit is non-linear!
§ NOT APPLICABLE on controlled sources!
§ NOT APPLICABLE on power contributions from distinct sources to
total power on resistors, total current passing resistor (or voltage
across it) must be calculated first (using superposition theorem) and
then power, P = RI 2 (power is not linear function)
Example – the circuit above, R1 = 100 W, R2 = 300 W, R3 = 200 W, U = 250 V, I
=1A
Ux = Ux1 + Ux2 = I
R2
R2R3
+U
= 120 + 150 = 270 V
R2 + R3
R2 + R3
,
P R2
2702 !
=
= 243 W
300
2
PU = U IU = U ( R2 U+R3 ¡ I R2R+R
)=
3
PI
300
= 250 ¢ ( 250
500 ¡ 1 ¢ 500 ) = 250 ¢ (¡0:1) = ¡25 W
h
i
R2 R3
= UI ¢ I = I ¢ (R1 + R2+R3 ) + Ux2 I =
= [1 ¢ (100 + 120) + 150] ¢ 1 = 370 W
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
- 10 -
Delta-star transformation
1
1
R01
R03
R12
R31
R02
2
3
3
R23
2
§ The properties of all terminal pairs must be the same in both delta (∆)
and star (wye, or Y) connections
§ In fact, total resistivity between distinct terminal pairs must be the same
in both connections:
R01 + R02 =
R12(R23 + R31)
R12 + R23 + R31
R02 + R03 =
R23(R31 + R12)
R12 + R23 + R31
R01 + R03 =
R31(R12 + R23)
R12 + R23 + R31
§ Then, ∆-Y transformation (replacement of ∆ connection by equivalent Y
connection) is
R01 =
R12R31
R12 + R23 + R31
R02 =
R12R23
R12 + R23 + R31
R01 =
R23R31
R12 + R23 + R31
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
- 11 -
§ Y-∆ transformation (replacement of Y connection by equivalent ∆
connection)
a) Total conductance between distinct terminal pairs must be the
same in both connections:
G03 (G01 + G02)
G31 + G23 =
G01 + G02 + G03
G12 + G23 =
G02 (G01 + G03)
G01 + G02 + G03
G12 + G31 =
G01 (G02 + G03)
G01 + G02 + G03
b) Then,
G12 =
G01G02
G01 + G02 + G03
G31 =
G01G03
G01 + G02 + G03
G23 =
G02G03
G01 + G02 + G03
c) Finally,
R12 = R01 + R02 +
R01R02
R03
R23 = R02 + R03 +
R02R03
R01
R31 = R03 + R01 +
R03R01
R02
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
- 12 -
Example:
R14
R4
R1
R3
R2
Ux = U
R14 =
R34
R13
Ux
In this circuit, resistors R1, R3 and R4
form ∆ connection. Computation of
voltage Ux is laborious using step by
step simplification or other methods.
R4
R3
R1
R2
Ux
Here, the resistors R1, R3 and R4 were
transformed from ∆ connection to Y
connection. Now, it is just simple
voltage divider.
R2 + R13
R14 + R13 + R2
R1R4
,
R1 + R3 + R4
R13 =
R1R3
,
R1 + R3 + R4
R34 =
R3R4
R1 + R3 + R4
Mnemonic: if in distinct node in ∆ connection are connected resistors Rx and
Ry, then is convenient denote resistor connected into same node in Y
connection Rxy. Then in nominator is product RxRy, in denominator sum of all
three resistors in ∆ connection.
Opposite procedure:
R14R13
R1 = R14 + R13 +
,
R34
R4 = R14 + R34 +
R3 = R13 + R34 +
R13R34
,
R14
R14R34
R13
Mnemonic: from Y to ∆ connection – if between two nodes are connected
resistors Rx andR
Ry in Y connection, then the resistivity of resistor, connected
R R
between same nodes is Rx + Ry + Rx z y .
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
- 13 -
Coupled and not coupled inductors
Sometimes it could be useful replace mutual coupled inductors by ordinary
uncoupled inductors. If coupled inductors are connected into same node, then
the replacement is:
Pavel Máša – ELECTRICAL CIRCUITS 1 – LECTURE 3:
ELEMENTARY ANALYSIS OF RESISTIVE CIRCUITS
- 14 -