Lecture 01�Circuit model and circuit law
Circuit and Circuit model
Circuit and Circuit model
Circuit and Circuit model
Circuit and Circuit model
Circuit and Circuit model
(a relationship of current and voltage)
not decomposable
Circuit and Circuit model
same main electromagnetic characteristics can be represented by the same circuit model under certain conditions (e.g., two resistors of 5 ohm and 10 ohm).
Inductance coil circuit model
Reference direction of current and voltage
Ref direction
actual direction
i >0
Ref direction
actual direction
i <0
Reference direction of current and voltage
Ref. direction
Ref. direction
і
A
B
A
B
Ref. direction
B
іAB
Reference direction of current and voltage
A
B
C
It is known that a particle with 4C positive charge moves uniformly from point A to point B, and the electric field force does 8 joules of work, and the electric field force does 12 joules of work from point B to point C.
φB = 0v, Since (φA-φB) x q = (φA-0) x 4 = 8, we have φA= 2v
Since (φB-φC) x q = (0-φC) x 4 = 12, we have φC= -3v
uAB = φA-φB =2v, uBC = φB-φC =3v
(2) If C is taken as the reference point, calculate the potential of points A, B and C, and the voltage of uAB and uBC
Try it yourself after class
Reference direction of current and voltage
A
Ref. direction
B
uAB >0
+
-
+
-
actual potential
A
Ref. direction
B
+
-
-
+
actual potential
uAB <0
A
B
A
B
+
-
arrow representation
pos. neg. sign representation
A
B
subscript representation
uAB
Reference direction of current and voltage
A
B
non-associated reference directions
A
B
+
-
associated reference directions
u
і
+
-
і
A
B
+
-
In terms of A and B, whether are they
associated reference directions?
Reference direction of current and voltage
Electric power and energy
P = dw/dt, u=dw/dq, i=dq/dt
P = dw/dt = dw/dq * dq/dt = u*i
A
B
+
-
associated reference directions
u
і
P = u*I to represent absorbing power
If P>0, absorbing power (actually)
If P<0, emitting power (actually)
an arbitrary
circuit component
A
B
-
+
non-associated reference directions
u
і
P = u*I to represent emitting power
If P>0, emitting power (actually)
If P<0, absorbing power (actually)
an arbitrary
circuit component
Example:
+
+
+
+
+
+
-
-
-
-
-
-
i1
i2
i3
u2
u3
u4
u1
u5
u6
u1=1v
u2=-3v
u3=8v
u4=-4v
u5=7v
u6=-3v
i1=2A
i2=1A
i3=-1A
positive values: the reference
direction is the same as actual
direction.
negative values: the reference
direction is the opposite of actual
direction.
p1=u1xi1=2w (emitting)
p2=u2xi1=-6w (emitting)
p3=u3xi1=16w (absorbing)
p4=u4xi2=-4w (emitting)
p5=u5xi3=-7w (emitting)
p6=u6xi3=3w (absorbing)
Circuit element (component)
Note: If the mathematical relationship characterizing the characteristics at both ends of the element is linear, the element is linear. Otherwise, it is a nonlinear element. E.g., resistor, inductive, and capacitance elements.
Circuit element (component)
Resistor element (component)
i
u
f(u,i)=0
R
GB (national standard)
R
widely adopted internationally
Resistor element (component)
R
i
+
-
u
u,i take associated reference direction
i
u
volt-ampere characteristic curve
Resistor element (component)
Resistor element (component)
Ohm‘s law
R
i
-
+
u
u,i take non-associated reference direction
Resistor element (component)
R
i
+
-
u
u,i take associated reference direction
p=u*i = u*u/R = i*R*R
R
i
-
+
u
u,i take non-associated reference direction
p=u*i = -u*u/R = -i*R*R
For both cases, a resistor element always absorbs power.
Resistor element (component)
R
i
+
-
u
u,i take associated reference direction
u
+
-
R
i
u
+
-
i
open circuit
i=0,u=?
R = ∞, G=0
u
+
-
R
i
u
+
i
short circuit
u=0,i=?
R=0,G= ∞
volt-ampere characteristic curve
-
u
i
u
i
Voltage and current sources
independent of the magnitude and direction (actual direction)
of the current flowing through the element and the external
circuit
i
-
+
us
GB (national standard)
i
-
+
us
widely adopted internationally
i
us
volt-ampere characteristic curve of ideal DC voltage source
Voltage and current sources
2. The current passing through the voltage source is jointly determined by the voltage source itself and the external circuit.
i=us/R
R = ∞, i=0
R = 0, i= ∞
+
-
R
i
us
Voltage source cannot be short circuited!
+
-
i
us
+
-
P=i*us>0,
emitting power
u,i take non-associated reference direction
Physical meaning: the current (positive charge) moves from low potential to high potential, voltage source overcomes the electric field force to work, and generates power.
i>0 and us>0,
i<0 and us<0,
Voltage and current sources
i
u,i take associated reference direction
+
-
us
P=i*us>0,
aborbing power
acting as a load
Physical meaning: the current (positive charge) moves from high potential to low potential, the electric field force works, and voltage source absorbs power.
i>0 and us>0,
i<0 and us<0,
+
-
us
+
-
us
R=5Ω
5v
10v
+
-
uR
uR = 10-5 = 5V
i = uR/R = 1A
PR = 5*1 = 5W (absorbing)
P10V = 10*1 = 10W (emitting)
P5V = 5*1 = 5W (absorbing)
Note: voltage source can
absorb power. For example, rechargeable batteries
Voltage and current sources
independent of the magnitude and direction (actual
direction) of the voltage at both end of it and the
external circuit
is
-
+
u
GB (national standard)
u,is can take either associated or non- associated reference direction
is
u
volt-ampere characteristic curve of ideal DC current source
is
widely adopted internationally
Voltage and current sources
2. The actual voltage at the two ends of the ideal current source is jointly determined by the current source itself and the external circuit.
is
-
+
u
R
u=R*is
R = ∞, u= ∞
R = 0, u= 0
Ideal current source cannot be open circuited!
Current transformer
Voltage and current sources
It can be generated by current stabilizing electronic devices. For example, the collector current of the transistor is independent of the load, and the photocell generates current under the irradiation of light, etc.
is
-
+
u
P=is*u>0,
emitting power,
acting as a power supply
u,is take non-associated reference direction
is
-
+
u
P=is*u>0,
absorbing power,
acting as a load
u,is take associated reference direction
Voltage and current sources
+
-
2A
us
is
-
+
5V
i
u
u=5V,
i=-2A
P2A = 5*2 = 10W (emitting power)
P5V = 5*(-2) = -10W (absorbing power)