### Inclass demo of Norton EC

```3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
R = 5[]
1
R3 =
33[]
vS1 =
18[V]
+
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
R5 =
12[]
R4 =
56[]
vS2 = 7[V]
+
1.
2.
3.
4.
-
If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit.
If not sure about dependent or independent sources, then for both Thevenin and Norton (a 1st or b 1st OK):
a. find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM
tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).
b. find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM;
MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes).
Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if
it is a completely different problem.
c. Obtain REQ= VOC / ISC. Go to 4.
For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all
the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just
Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
R = 5[]
1
R3 =
33[]
vS1 =
18[V]
+
-
R4 =
56[]
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
-
+
B
A
R5 =
12[]
vS2 = 7[V]
+
1.
2.
3.
4.
-
If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit.
If not sure about dependent or independent sources, then for both Thevenin and Norton (a 1st or b 1st OK):
a. find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM
tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).
b. find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM;
MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes).
Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if
it is a completely different problem.
c. Obtain REQ= VOC / ISC. Go to 4.
For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all
the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just
Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
R = 5[]
1
R3 =
33[]
vS1 =
18[V]
+
-
R4 =
56[]
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
-
+
B
A
R5 =
12[]
vS2 = 7[V]
+
1.
2.
3.
4.
-
If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit.
If not sure about dependent or independent sources, then for both Thevenin and Norton (a 1st or b 1st OK):
a. find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM
tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).
b. find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM;
MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes).
Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if
it is a completely different problem.
c. Obtain REQ= VOC / ISC. Go to 4.
For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all
the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just
Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
2.
If not sure about dependent or independent
sources, then for both Thevenin and Norton:
a. find open circuit voltage VOC: the voltage across
the 2 points of interest. Use KVL, NVM, or
MCM; NVM tends to give voltage directly,
(which might be a bit more convenient, but
depending on the problem).
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached to
a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides node
voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
C
2.
If not sure about dependent or independent
sources, then for both Thevenin and Norton:
a. find open circuit voltage VOC: the voltage across
the 2 points of interest. Use KVL, NVM, or
MCM; NVM tends to give voltage directly,
(which might be a bit more convenient, but
depending on the problem).
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached to
a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides node
voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
C
2.
If not sure about dependent or independent
sources, then for both Thevenin and Norton:
a. find open circuit voltage VOC: the voltage across
the 2 points of interest. Use KVL, NVM, or
MCM; NVM tends to give voltage directly,
(which might be a bit more convenient, but
depending on the problem).
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
C
2.
R1i1
 vS 1
i1
R2i1
R5i1
If not sure about dependent or independent
sources, then for both Thevenin and Norton:
a. find open circuit voltage VOC: the voltage across
the 2 points of interest. Use KVL, NVM, or
MCM; NVM tends to give voltage directly,
(which might be a bit more convenient, but
depending on the problem).
 vS 2
i1 R1  R2  R5   vS 1  vS 2  0
v A  R5i1  vS 2
vS1  R1i1  R2i1  v A
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
R = 5[]
1
R3 =
33[]
vS1 =
18[V]
+
-
R4 =
56[]
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
I SC
-
B
+
A
R5 =
12[]
vS2 = 7[V]
+
1.
2.
3.
4.
-
If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit.
If not sure about dependent or independent sources, then for both Thevenin and Norton (a 1st or b 1st OK):
a. find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM
tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).
b. find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM;
MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes).
Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if
it is a completely different problem.
c. Obtain REQ= VOC / ISC. Go to 4.
For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all
the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just
Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
C
2. If not sure about dependent or independent
sources, then for both Thevenin and Norton:
b. find short circuit ISC: connect a wire across
the 2 points and find its current. Use KCL,
NVM or MCM; MCM tends to give current
directly, but may NOT be convenient if too
many equations (many meshes). Most
critical: this circuit has NOTHING to do with
circuit in #2a. It is a different circuit and
solve it as if it is a completely different
problem.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
2. If not sure about dependent or independent
sources, then for both Thevenin and Norton:
c. Obtain REQ= VOC / ISC. Go to 4.
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
2. If not sure about dependent or independent
sources, then for both Thevenin and Norton:
c. Obtain REQ= VOC / ISC. Go to 4.
v A  vS 1 0  vS 1

R2  R1 R2  R1
v A  ( vS 2 ) 0  vS 2

R5
R5
iSC 
vS 1
v
 S 2  0
R2  R1 R5
3.
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
4.
Draw TEQ or NEQ with proper source with
correct polarity and equivalent resistor.
A
INo
iS 2  0.3 A
RNo
iRNo
B
a) Find the Norton equivalent seen by the 0.3 A current source.
b) Find the power delivered by the 0.3 A current source.
R = 5[]
1
R3 =
33[]
+
3.
vS1 =
18[V]
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
R5 =
12[]
R4 =
56[]
vS2 = 7[V]
+
-
Can we do an independent verification of
the power (delivered or absorbed) by the
0.3 A source?
3.
b) Find the power delivered by the 0.3 A current source.
R1 = 5[]
R3 =
33[]
vS1 =
18[V]
+
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
R5 =
12[]
R4 =
56[]
vS2 = 7[V]
+
-
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached to
a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides node
voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
3.
b) Find the power delivered by the 0.3 A current source.
C
R1 = 5[]
R3 =
33[]
vS1 =
18[V]
+
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
A
R4 =
56[]
B
R5 =
12[]
vS2 = 7[V]
+
-
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached to
a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides node
voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
3.
b) Find the power delivered by the 0.3 A current source.
C
R1 = 5[]
R3 =
33[]
vS1 =
18[V]
+
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
A
R4 =
56[]
B
R5 =
12[]
vS2 = 7[V]
+
-
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached
to a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides node
voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
3.
b) Find the power delivered by the 0.3 A current source.
C
R1 = 5[]
R3 =
33[]
vS1 =
18[V]
+
-
R2 =
9[]
iS1 =
15[A]
iS2 = 0.3[A]
A
R4 =
56[]
B
R5 =
12[]
vS2 = 7[V]
+
-
1. Identify the node, choose a reference.
2. Identify node voltage that is already known (given)
3. Apply KCL to each unknown node. If no voltage source directly attached to
a node:
a. Draw current vector away from node
b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R
c. If a current is to a current source, write the current source with proper
polarity
d. Add all the currents and let = 0
4. If a direct voltage source attached to a node:
a. An equation can be: VX-VY=VS where VY is the node at the other side of
the voltage source.
b. If the source is in series with a resistor the other terminal (and no
branching), Norton equivalent circuit can be applied; or
c. An unknown current can be introduced to be solved later.
5. Assemble all the equations and identified additional unknown besides
node voltage.
6. Solve the equations
7. Use the known node voltage to derive other quantities asked by the
problem
C
R1 = 5[]
R3 =
33[]
vS1 =
18[V]
R2 =
9[]
iS1 =
15[A]
+
-
iS2 = 0.3[A]
A
B
R4 =
56[]
R5 =
12[]
vS2 = 7[V]
+
-
R1 , R2 , R3 , R5 , vS1 , vS2 , iS2
Solve
vA
vA
vS1
R1
R2
iS2
vA
5., 9 , 33 , 12., 18, 7., 0.3 ;
vS2
R5
0 , vA
2.6
Power delivered (active convention) by the 0.3 A source:
P
v A iS2 . v A
0.78
2.6
```