Circuit Elements
Resistor v = Ri
Voltage Source v = Vs
Current Source i = Is
dv
Capacitor i = C
dt
di
Inductor v = L
dt
Kirchhoff’s Laws
A node is a point where two or more
elements are connected.
A loop is a closed path through the circuit.
Kirchhoff’s Current Law (KCL):
The sum of all currents at a node, taken
with the appropriate sign, is zero.
Kirchhoff’s Voltage Law (KVL):
The sum of all voltages around a loop,
taken with the appropriate sign, is zero.
Polarity Conventions
Passive Convention
If vi > 0 then . . .
Power is dissipated in element.
Power flows from circuit to element.
If vi < 0 then . . .
Power is generated in element.
Power flows from element to circuit.
Ohm’s Law is v = +Ri
Active Convention
If vi > 0 then . . .
Power is generated in element.
Power flows from element to circuit.
If vi < 0 then . . .
Power is dissipated in element.
Power flows from circuit to element.
Ohm’s Law is v = −Ri
We could agree to . . .
(A) Allow either convention with any
element type.
(B) Always use passive convention with
RLC elements.
Always use active convention with sources.
(C) Always use passive convention, with
every circuit element.
(C) has several advantages . . .
1. Always get a + in element equations.
2. All elements are treated in the same
way.
3. vi > 0 always means that energy is
being dissipated in the element, no
matter what type of element it is.
I will adhere to (C)
The quadrants of an element’s v − i
characteristic correspond to electrical
energy being generated or dissipated in the
element.
For the passive convention . . .
6
Generation
Dissipation
-
Dissipation
Generation
For the passive convention . . .
6
Generation
Dissipation
-
Dissipation
Generation
The situation is reversed with the active
convention.
Resisitors always dissipate energy.
Voltage and Current Sources may either
generate or dissipate energy.
The Direct Method
Uses the direct application of the element
equations and Kirchhoff’s Laws.
Step 1: Label the voltage across each
element.
Step 2: Label the current through each
element.
Step 3: Write the element equation for
each element.
Step 4: Use KCL to write an equation for
each node.
Step 5: Use KVL to write an equation for
certain loops.
Step 6: Solve the resulting set of equations.
Example
6
Find the voltage across the 6Ω resistor.
5 Element Equations . . .
v1 = 2
v2 = 4i2
v3 = 5i3
v4 = 6i4
i5 = 3
3 Nodes (KCL)
A i1 + i 2 = 0
B −i2 + i3 + i4 = 0
C −i3 − i5 = 0
2 Loops (KVL)
D v2 + v4 − v1 = 0
E v4 − v3 + v5 = 0
Have 10 equations in 10 unknowns
Solve equations for solution
Example
Find the voltage across and the current
through every element, and the power
dissipated or generated in every element.
5 Element Equations
i1 = 5
v2 = 2i2
v3 = 3i3
v4 = 4i4
v5 = 8
3 Nodes (KCL)
A i1 = i2
B i2 = i3 + i 4
C i4 = i5
2 Loops (KVL)
D v1 + v2 + v3 = 0
E v4 + v5 = v3
Have 10 equations in 10 unknowns
Use element equations to eliminate
i1 , v 2 , v 3 , v 4 , v 5
A i2 = 5
B i2 = i3 + i4
C i4 = i5
D v1 + 2i2 + 3i3 = 0
E 4i4 + 8 = 3i3
Use A to eliminate i2 = 5
B i3 + i4 = 5
C i4 = i5
D v1 + 3i3 = −10
E 4i4 + 8 = 3i3
Use B to eliminate i3 = 5 − i4
C i4 = i5
D v1 − 3i4 = −25
E 4i4 + 8 = 15 − 3i4
E now gives 7i4 = 7 ⇒ i4 = 1
D ⇒ v1 = −22
C ⇒ i4 = i5 = 1
B ⇒ i3 = 4
A ⇒ i2 = 5
Element equations then give . . .
i1 = 5, v2 = 10, v3 = 12, v4 = 4, v5 = 8
Power Dissipated in each element . . .
v2i2 = 50W
v3i3 = 48W
v4i4 = 4W
v1i1 = −110W dissipated = +110W
generated
v5i5 = +8W
The voltage source absorbs energy,
“recharging”
Law of conservation of energy . . .
50+48+4+8=110
Loop Current Method
For circuits with no current sources, the
procedure is . . .
Step 1. Select the minimum number of
loops so that every circuit element is part
of at least one loop.
Step 2. Label the current flowing around
each loop, and its direction.
Step 3. Apply KVL and the element
equations simultaneously to each loop.
Step 4. Solve the resulting set of linear
equations.
Example
2Ω
3Ω
'$
1V
&%
'$
4Ω
5V
&%
Find the voltage across and current
through each resistor.
Loop 1. 2i1 + 4(i1 − i2) − 1 = 0
⇒ 6i1 − 4i2 = 1
Loop 2. 3i2 + 4(i2 − i1) + 5 = 0
⇒ 7i2 − 4i1 = −5
Solving gives i2 = −1, i1 = −0.5
2Ω
3Ω
'$
1V
&%
'$
4Ω
5V
&%
Both (i) KVL and (ii) the element
equations are applied when writting down
the equations.
KCL is automatically obeyed because of
the way in which the currents are labelled.
Consider what happens at a node.
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When writing down the equation for ik ’s
loop, terms involving ik always have the
same sign when written on the same side of
the equals sign.
For circuits with one or more current
sources, one need to, in addition,
1. Label the voltage across each current
source.
2. Write an equation for each current
source.
Example
4Ω
3Ω
'$
'$
5A &%
&%
Loop 1:
Loop 2:
I-source:
3i1 − v1 = 0
v1 + 4i2 + 6 = 0
5 = i2 − i1
6V
(1)
(2)
(3)
(1) ⇒ v1 = 3i1
(2) ⇒ 3i1 + 4i2 = −6
4i2 = −6 − 3i1 = 20 + 4i1
⇒ 7i1 = −26
9
⇒ i 2 = 5 + i1 =
7