Use Nodal Analysis to solve for all of the branch currents and node voltages in the circuits below. Show
all work.
Note that the solutions for the node voltage depend on which node you chose as ground. However,
magnitude of the voltage drops and the currents will be unaffected by this choice.
1.
I3
I1
Vd
Vb
Va
Ibat
Ve
I5
Vc
I6
I4
I2
@𝑉𝑎 : − 2m + I1 = 0
−2m +
@𝑉𝑏 : − I1 + I3 + 2m = 0
Vb −Va
5k
+
Va − Vb
=0
5k
Vb −Vd
6k
(1)
+ 2m = 0 (2)
By substituting the first equation into the second, we get:
−2m +
Thus, I3 = 0 A.
Vb − Vd
+ 2m = 0 or simply: Vb = Vd
6k
@𝑉𝑐 : 2m − 2m + I2 = 0
2m − 2m +
Vc
= 0 or Vc = 0 V (3)
3k
@𝑉𝑑 : −Ibat −I3 + I5 + I6 = 0
Vd − Vb Vd − Ve Vd
−Ibat +
+
+
=0
6k
1k
2k
Notice that we cannot directly express 𝐼𝑏𝑎𝑡 in terms of node voltages. We will return to this shortly.
@𝑉𝑒 : Ibat +I4 − I5 = 0
Ibat +
Ve Ve − Vd
+
=0
8k
1k
By plugging this equation into the previous one (@Vd) :
Ve Ve − Vd Vd − Vb Vd − Ve Vd
+
+
+
+
=0
8k
1k
6k
1k
2k
Noting that Vb = Vd (from the equation @Vb), the above equation simplifies to:
𝑉𝑒 𝑉𝑑
+
= 0 (4)
8𝑘 2𝑘
Since the two equations established at Vd and Ve collapsed into one as a result of Ibat , we require an
additional equation to solve for the 5 unknowns. We obtain this equation by noting that:
Vd − Ve = 10V
(5)
From (4) and (5), we find Ve = −8 and V𝑑 = 2.
Thus far, we have:
Vb = 2 V, 𝑉𝑐 = 0 𝑉, Vd = 2 V, and Ve = −8 V
Finally, we find 𝑉𝑎 using the equation @𝑉𝑎 :
Va = 2m(5k) + Vb = 12 𝑉
Now we can find all of the currents from:
Va=12V
I3=0 A
I1=2 mA
Vb=2 V
Vd=2 V
Ibat=11 mA
Vc=0V
Ve=-8V
𝑉𝑎 − 𝑉𝑏
,
5𝑘
I6=1 mA
I4=-1 mA
I2=0 A
𝐼1 =
I5=10 mA
𝐼2 =
𝑉𝑐
𝑉𝑒
10
𝑉𝑑
, 𝐼3 = 0 𝐴, 𝐼4 =
, 𝐼5 =
, 𝐼6 =
, and 𝐼𝑏𝑎𝑡 = 𝐼5 − 𝐼4
3𝑘
8𝑘
1𝑘
2𝑘
Voltage
12 V
Va
2V
Vb
0V
Vc
2V
Vd
-8
V
Ve
I1
I2
I3
I4
I5
I6
Ibat
Current
2 mA
0A
0A
-1 mA
10 mA
1 mA
11 mA
Note that the voltage drop Va-Vc = 12 V while the voltage drop Vc-Vb = -2 V. This means that the
current source on the left is generating power (- 24 mW). However, the current source on the right is
dissipating power (+4 mW). Just because the circuit has three power supplies in it does not mean that all
of the power supplies are generating power. Power supplies may dissipate power as well as generate it.
For example, a car battery is a d.c. voltage supply that generates power when you use it to start your car’s
engine. However, it dissipates power as the alternator charges the battery while you drive.
I4
I2
I1
I7
Vd=-5 V
2.
I6
Vc
I3
Vb
I5
By selecting the node below the 5 V source as ground, we can easily determine Vd as -5 V or:
0 − Vd = 5 → Vd = −5 V
Furthermore, we can solve for 𝑉𝑎 by noting that:
Vd − 𝑉𝑎 = 3 → −5 − 𝑉𝑎 = 3 or 𝑉𝑎 = −8 V
Now, we are left with only 2 unknowns (Vb and Vc), and therefore 2 equations.
@𝑉𝑏 :
− 𝐼5 + 𝐼6 − 𝐼7 = 0
Vb − Vc Vb + 5 Vb + 8
Vb Vb Vb − Vc
+
+
= 0 or
+
+
= −9.25 m
2k
4k
1k
1k 4k
2k
@Vc : I3 + I4 + I5 − 1m = 0
Vc
Vc + 5 Vc − Vb
Vc − Vb Vc
Vc
+
+
− 1m = 0 or
+
+
= −4 m
10k
1k
2k
2k
1k 10k
Solving the two equations yields: 𝑉𝑏 = −6.59 𝑉 and 𝑉𝑐 = −4.56 𝑉
So we have:
𝑉𝑎 = −8 V ,
𝑉𝑏 = −6.59 𝑉 and 𝑉𝑐 = −4.56 𝑉
The currents are then:
I1 =
I4 =
5
= 1.67 mA,
3k
𝑉𝑐 + 5
= 0.44 mA,
1k
𝐼2 = 𝐼3 − 𝐼1 = −2.12 mA I3 =
I5 =
Vc
= −0.456 mA
10k
Vc − 𝑉𝑏
𝑉𝑏 + 5
= 1.02 mA 𝐼6 =
= −0.397 mA,
2k
4𝑘
Va
𝐼7 =
𝑉𝑎 − 𝑉𝑏
= −1.41 mA
1𝑘
In summary:
Voltage (V)
-8
Va
-6.59
Vb
-4.56
Vc
(Vd=-5 V)
Current(mA)
I1
1.67
I2
-2.12
I3
-0.456
I4
0.44
I5
1.02
I6
-0.397
I7
-1.41
I7=-1.41 mA
-5 V
-8 V
I4=0.44 mA
I1=1.67mA
I2=-2.12 mA
-4.56 V
I3=-0.46 mA
I6=
-0.397 mA
-6.59 V
I5=1.02 mA
3.
𝐼1
𝑉𝑏
𝐼7
𝑉𝑎
𝐼1
𝑉𝑒
𝐼5
𝐼3
𝐼2
0 𝑉 (𝐺𝑁𝐷)
𝐼4
First, we can write by inspection:
Va − Vb = 8 V
Vd − Vc = 20 V
(1)
(2)
Now, we apply KCL at each node:
@𝑉𝑎 : I1 + I2 − I1 + I7 + I3 − I5 = 0
Since 𝐼1 cancels out:
I2 + I7 + I3 − I5 = 0
𝑉𝑎 𝑉𝑎 − 𝑉𝑐 𝑉𝑎 𝑉𝑎 − 𝑉𝑒
+
+ +
=0
10
15
5
3
Which simplifies to:
(
1 1 1 1
−𝑉𝑐 −𝑉𝑒
+ + + ) 𝑉𝑎 +
+
= 0 (3)
10 5 3 15
15
3
Next:
@𝑉𝑏 : −𝐼1 + 𝐼1 = 0 or −𝐼1 +
𝑉𝑏 −𝑉𝑎
2
𝐼7
𝑉𝑐
= 0 → 𝐼1 =
𝐼1 =
𝑉𝑏 −𝑉𝑎
2
𝑉𝑏 − 𝑉𝑎
2
(4)
𝐼6
𝑉𝐷
@𝑉𝑐 : − 𝐼7 + 𝐼7 = 0 or
𝐼7 =
𝑉𝑐 −𝑉𝑎
15
+ 𝐼7 = 0 → 𝐼7 =
𝑉𝑎 −𝑉𝑐
15
𝑉𝑎 − 𝑉𝑐
15
@𝑉𝑑 :
− 𝐼7 + 𝐼6 + 𝐼4 + 1 = 0
−𝐼7 +
𝑉𝑑 − 𝑉𝑒 𝑉𝑑
+ +1=0
2
3
We can substitute in 𝐼7 =
𝑉𝑎 −𝑉𝑐
15
−
from the previous equation and rearrange to get:
1
1
1 1
1
𝑉𝑎 + 𝑉𝑐 + ( + ) 𝑉𝑑 − 𝑉𝑒 = −1 (5)
15
15
2 3
2
Finally,
@𝑉𝑒 : 𝐼5 − 𝐼6 − 1 = 0
𝑉𝑒 − 𝑉𝑎
𝑉𝑒 − 𝑉𝑑
+
−1=0
3
2
Which can be expressed as:
1
1
1 1
− 𝑉𝑎 − 𝑉𝑑 + ( + ) 𝑉𝑒 = 1
3
2
2 3
From nodal analysis, we have found 5 equations and 5 unknowns that are summarized below:
(1) Va − Vb = 8 V
(2) − Vc + Vd = 20 V
1 1 1 1
−𝑉𝑐 −𝑉𝑒
(3) ( + + + ) 𝑉𝑎 +
+
=0
10 5 3 15
15
3
𝑉𝑏 − 𝑉𝑎
(4) 𝐼1 =
(𝑛𝑜𝑡 𝑢𝑠𝑒𝑓𝑢𝑙)
2
1
1
1 1
1
(5) −
𝑉𝑎 + 𝑉𝑐 + ( + ) 𝑉𝑑 − 𝑉𝑒 = −1
15
15
2 3
2
1
1
1 1
(6) − 𝑉𝑎 − 𝑉𝑑 + ( + ) 𝑉𝑒 = 1
3
2
2 3
Notice that the nodal equations yielded 6 equations, but (4) is not useful as it contains 𝐼1 . As a result, we
are left with 5 independent equations, which is adequate to find the 5 nodal voltages.
Solving the set of equations yields:
Voltage
-1.16 V
𝐕𝐀
-9.16 V
𝐕𝐁
-18.96 V
𝐕𝐂
1.04 V
𝐕𝐃
1.36 V
𝐕𝐄
𝑉𝑏 =
−9.16 𝑉
I1
I2
I3
I4
I5
I6
I7
𝐼7 =
1.19 A
𝐼1 = −4 A
Current
-4 A
-116 mA
-231 mA
347 mA
840 mA
-160 mA
1.19 A
𝑉𝑐 =
−18.96 𝑉
𝑉𝑑 = 1.04 𝑉
𝑉𝑒 = 1.36 𝑉
𝑉𝑎 = −1.16 𝑉
𝐼3 =
−0.231 A
𝐼5 =
0.84 A
𝐼2 = −0.116 A
0 𝑉 (𝐺𝑁𝐷)
𝐼4 =
0.347 A
𝐼6 =
−0.16 A