ENGG 1300
,ENGG 1300 Notes :
Nodal
Analysis :
1.)
Votage at ground node ou -
-
2.) Equations for voltage and current sources .
8.) Equations for Resistors
KCL at all nodes
4.)
Apply
5.) Eliminate current from equations and solve for node voltages
6.) solve for branch currents
7.) If needed solve for branch voltages and powers
,
ioosr
List of Unknowns
JENNY VA -
-
IOV Ii = ?
VB = ? Iz= ?
look 2005L
Ve ? Is
•MhÉ•B→tk•o
-
= ?
A
-
1¥ fits
.
ft Is " VD= 0 Iy= 0.02 A
§ ⑧ 20mA 3300 Ig-
"
IOV ? =
1-D- Is ? =
DE
*
steps Voltage at ground node ou
: -
-
*
step 2 componentIOVequations for voltage
: a current sources
Ltg VA -
VgOV -
-
VD
i. -
-
f VA = IOV
LDA I4= 20mA = 20×10-3A
-50.02A
,* Step 3: component equations for resistors
lie relate node for
.
voltages to branch current
yeah resistor)
↳ggIz=VA-pfB_ =V%%→
A
EM look
HlbÉ•-MN-•0
gg
zoon
VA =/ OV
10,%÷
,
¥2 =
tts ft Is Is
=%¥- __VB¥•_o
" Is "
Boor ⑧ 20mA
.
lol
y ↳•£5=
¥É- 1-4=0.02 A ¥0
Io=V%oVe_ Is
10,2¥
-
-
* step 4 Apply: Nodal Analysis
1Wh @ Nodes
LDK Node (A) : I, t Iat Io -0 ①
48 Node (B) ÷
Iz + Is =
Is ②
LDK Node ¢03 : Ia 1- Is =
Is ③
* steps : Eliminate currents from nodal equations
( substitute component
LA ① I , + Izt Is -0
equations into KVL )
Ii
t.IO#B-tloo-oVc---
0
②
10-VB-tvo-VB-Vg.gg
100 200
③ 0.02 t 0T¥
'
=
VIo¥-
, Solve
* equations for Vrs and Vo
① It
1%VB_t%Ve_
= 0
,%Itv%V;_=§÷
1
÷→÷ Yoo :%=¥o
- -
÷
1%+2%-0 3%4%+2%0
=
i. I
oB_
o.lt =
200
÷÷÷÷.÷÷÷÷
② 60 t 3 Vc = HUB
0.02 t
%%-=VIo¥-
'
24+38-4=4
•• i. Vc= 10.8 Volts
012+2%-0=1%0 ¥0 +
: .
Is =
10.8-8.4-10.012 A
0-12 t
Viggo 3%0 200
=
I5=
8g¥
i. = 0.028 A
③ 24 t VB =3 Vc
Is
1,0010¥
i. = = -0.008A
i. 60 + (24 + VB) =
11 VB
84 = 10 VB ÷I' =
-12-16=-0.008 *
i. VB = 8.4 Volts
,ENGG 1300 Notes :
Nodal
Analysis :
1.)
Votage at ground node ou -
-
2.) Equations for voltage and current sources .
8.) Equations for Resistors
KCL at all nodes
4.)
Apply
5.) Eliminate current from equations and solve for node voltages
6.) solve for branch currents
7.) If needed solve for branch voltages and powers
,
ioosr
List of Unknowns
JENNY VA -
-
IOV Ii = ?
VB = ? Iz= ?
look 2005L
Ve ? Is
•MhÉ•B→tk•o
-
= ?
A
-
1¥ fits
.
ft Is " VD= 0 Iy= 0.02 A
§ ⑧ 20mA 3300 Ig-
"
IOV ? =
1-D- Is ? =
DE
*
steps Voltage at ground node ou
: -
-
*
step 2 componentIOVequations for voltage
: a current sources
Ltg VA -
VgOV -
-
VD
i. -
-
f VA = IOV
LDA I4= 20mA = 20×10-3A
-50.02A
,* Step 3: component equations for resistors
lie relate node for
.
voltages to branch current
yeah resistor)
↳ggIz=VA-pfB_ =V%%→
A
EM look
HlbÉ•-MN-•0
gg
zoon
VA =/ OV
10,%÷
,
¥2 =
tts ft Is Is
=%¥- __VB¥•_o
" Is "
Boor ⑧ 20mA
.
lol
y ↳•£5=
¥É- 1-4=0.02 A ¥0
Io=V%oVe_ Is
10,2¥
-
-
* step 4 Apply: Nodal Analysis
1Wh @ Nodes
LDK Node (A) : I, t Iat Io -0 ①
48 Node (B) ÷
Iz + Is =
Is ②
LDK Node ¢03 : Ia 1- Is =
Is ③
* steps : Eliminate currents from nodal equations
( substitute component
LA ① I , + Izt Is -0
equations into KVL )
Ii
t.IO#B-tloo-oVc---
0
②
10-VB-tvo-VB-Vg.gg
100 200
③ 0.02 t 0T¥
'
=
VIo¥-
, Solve
* equations for Vrs and Vo
① It
1%VB_t%Ve_
= 0
,%Itv%V;_=§÷
1
÷→÷ Yoo :%=¥o
- -
÷
1%+2%-0 3%4%+2%0
=
i. I
oB_
o.lt =
200
÷÷÷÷.÷÷÷÷
② 60 t 3 Vc = HUB
0.02 t
%%-=VIo¥-
'
24+38-4=4
•• i. Vc= 10.8 Volts
012+2%-0=1%0 ¥0 +
: .
Is =
10.8-8.4-10.012 A
0-12 t
Viggo 3%0 200
=
I5=
8g¥
i. = 0.028 A
③ 24 t VB =3 Vc
Is
1,0010¥
i. = = -0.008A
i. 60 + (24 + VB) =
11 VB
84 = 10 VB ÷I' =
-12-16=-0.008 *
i. VB = 8.4 Volts
