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.

@Va: -2m+I1=0

-2m+Va-Vb5k=0 (1)

@Vb: -I1+I3+2m=0

Vb-Va5k+Vb-Vd6k+2m=0 (2)

By substituting the first equation into the second, we get:

-2m+Vb-Vd6k+2m=0 or simply: Vb=Vd

Thus, I3=0 A.

@Vc: 2m-2m+I2=0

2m-2m+Vc3k=0 or Vc=0 V (3)

@Vd: -Ibat-I3+I5+I6=0

-Ibat+Vd-Vb6k+Vd-Ve1k+Vd2k=0

Notice that we cannot directly express Ibat in terms of node voltages. We will return to this shortly.

@Ve: Ibat+I4-I5=0

Ibat+Ve8k+Ve-Vd1k=0

By plugging this equation into the previous one (@Vd) :

Ve8k+Ve-Vd1k+Vd-Vb6k+Vd-Ve1k+Vd2k=0

Noting that Vb=Vd (from the equation @Vb), the above equation simplifies to:

Ve8k+Vd2k=0 (4)

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=2m5k+Vb=12 V

Now we can find all of the currents from:

I1=Va-Vb5k, I2=Vc3k, I3=0 A, I4=Ve8k, I5=101k, I6=Vd2k, and Ibat=I5-I4

Voltage / Current
Va / 12 V / I1 / 2 mA
Vb / 2 V / I2 / 0 A
Vc / 0 V / I3 / 0 A
Vd / 2 V / I4 / -1 mA
Ve / -8 V / I5 / 10 mA
I6 / 1 mA
Ibat / 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.

2.

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-Vc2k+Vb+54k+Vb+81k=0 or Vb1k+Vb4k+Vb-Vc2k=-9.25 m

@Vc: I3+I4+I5-1m=0

Vc10k+Vc+51k+ Vc-Vb2k-1m=0 or Vc-Vb2k+Vc1k+Vc10k=-4 m

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=53k=1.67 mA, I2=I3-I1=-2.12 mA I3=Vc10k=-0.456 mA

I4=Vc+51k=0.44 mA, I5=Vc-Vb2k=1.02 mA I6=Vb+54k=-0.397 mA,

I7=Va-Vb1k=-1.41 mA

In summary:

Voltage (V) / Current(mA)
Va / -8 / I1 / 1.67
Vb / -6.59 / I2 / -2.12
Vc / -4.56 / I3 / -0.456
(Vd=-5 V) / I4 / 0.44
I5 / 1.02
I6 / -0.397
I7 / -1.41

3.

First, we can write by inspection:

Va-Vb=8 V (1)

Vd-Vc=20 V (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

Va10+Va-Vc15+Va5+Va-Ve3=0

Which simplifies to:

110+15+13+115Va+-Vc15+-Ve3=0 (3)

Next:

@Vb: -I1+I1=0 or -I1+Vb-Va2=0→I1=Vb-Va2

I1=Vb-Va2 (4)

@Vc: -I7+I7=0 or Vc-Va15+I7=0→I7=Va-Vc15

I7=Va-Vc15

@Vd: -I7+I6+I4+1=0

-I7+Vd-Ve2+Vd3+1=0

We can substitute in I7=Va-Vc15 from the previous equation and rearrange to get:

- 115Va+115Vc+12+13Vd-12Ve=-1 (5)

Finally,

@Ve: I5-I6-1=0

Ve-Va3+ Ve-Vd2-1=0

Which can be expressed as:

-13Va-12Vd+12+13Ve=1

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

3 110+15+13+115Va+-Vc15+-Ve3=0

4 I1=Vb-Va2 (not useful)

5 - 115Va+115Vc+12+13Vd-12Ve=-1

6 -13Va-12Vd+12+13Ve=1

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 / Current
VA / -1.16 V / I1 / -4 A
VB / -9.16 V / I2 / -116 mA
VC / -18.96 V / I3 / -231 mA
VD / 1.04 V / I4 / 347 mA
VE / 1.36 V / I5 / 840 mA
I6 / -160 mA
I7 / 1.19 A