Mesh and Node Equations: Simple Resistive Circuits
Introduction
The circuits in this set of problems are small circuits that consist of independent sources,
resistors and a meter. In particular, these circuits do not contain dependent sources. These
circuits can be analyzed using mesh equation or using node equations.
Node equations are discussed in Sections 4.3 and 4.4 of Introduction to Electric Circuits by R.C.
Dorf and J.A Svoboda. Mesh equations are discussed in Section 4.6.
Worked Examples
Example 1:
Consider the circuit shown in Figure 1. Find the value of the current source current, Ia.
Figure 1 The circuit considered in Example 1.
Solution: Figure 2 shows the circuit from Figure 1 after replacing the ammeter by an equivalent
short circuit and labeling the current measured by the ammeter. This circuit can be analyzed
using mesh equations or using node equations. We will do both.
Figure 2 The circuit from Figure 1 after replacing the ammeter by a short circuit.
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Figure 3 The circuit from Figure 2 after labeling the mesh currents.
First, consider analyzing the circuit in Figure 2 using mesh equations. Figure 3 shows the
circuit after labeling the mesh currents, i1 and i2. The mesh current i2 is equal to the current in the
short circuit that replaced the ammeter so
i2 = 3 A
The current in the current source can be expressed in terms of the mesh currents as
I a = i2 − i1 = 3 − i1
(1)
Apply KVL to the supermesh to get
2 i1 + ( 2 )( 3) − 4 = 0 ⇒ i1 = −1 A
Substituting this value into Equation 1 gives I a = 3 − (−1) = 4 A .
Next, consider analyzing the circuit in Figure 2 using node equations. Figure 4 shows the
circuit after selecting a reference node and numbering the other nodes. Let v1 and v2 denote the
node voltages at node 1 and node 2, respectively.
The voltage of the voltage source can be expressed in terms of the node voltages as
4 = v1 − 0 ⇒ v1 = 4 V
The current in the short circuit that replaced the ammeter is equal to the current in one of the 2 Ω
resistors. This current can be expressed in terms of the node voltages as
v2 − 0
= 3 ⇒ v2 = 6 V
2
Apply KCL at node 2 to get
Ia =
v2 − v1
6−4
+3=
+3= 4 A
2
2
2
This value of current source current agrees with the value calculated using mesh equations, as it
must.
Figure 4 The circuit from Figure 2 after labeling the nodes. (The encircled numbers are node
numbers.)
Example 2:
Consider the circuit shown in Figure 5. Find the value of the current source current, Ia.
Figure 5 The circuit considered in Example 2.
Solution: Figure 6 shows the circuit from Figure 5 after replacing the voltmeter by an equivalent
open circuit and labeling the voltage measured by the voltmeter. This circuit can be analyzed
using mesh equations or using node equations. We will do both.
First, consider analyzing the circuit in Figure 6 using mesh equations. Figure 7 shows the
circuit after labeling the mesh currents, i1 and i2.
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Figure 6 The circuit from Figure 5 after replacing the voltmeter by a open circuit.
Figure 7 The circuit from Figure 6 after labeling the meshes.
The mesh current i2 is equal to the current in the 2 Ω resistor at the right of the circuit. The
voltmeter measured to voltage across that 2 Ω resistor to be 9 V. Using Ohm’s Law, the mesh
current is given by
9
i2 = = 4.5 A
2
The current in the current source can be expressed in terms of the mesh currents as
I a = i2 − i1 = 4.5 − i1
(2)
Apply KVL to the super mesh to get
8 i1 + ( 2 )( 4.5 ) + 9 − 6 = 0 ⇒ i1 = −1.5 A
Substituting this value into Equation 2 gives I a = 4.5 − (−1.5) = 6 A .
Next, consider analyzing the circuit in Figure 6 using node equations. Figure 8 shows the
circuit after selecting a reference node and numbering the other nodes. Let v1, v2 and v3 denote
the node voltages at nodes 1, 2 and 3, respectively.
The voltage of the voltage source can be expressed in terms of the node voltages as
6 = v1 − 0 ⇒ v1 = 6 V
4
The voltmeter measures the voltage at node 3 to be 9 V. That is v3 = 9 V . Apply KCL at node 3
to get
v3 − v2 v3
9 − v2 9
+ =0 ⇒
+ = 0 ⇒ v2 = 18 V
2
2
2
2
Apply KCL at node 2 to get
Ia =
v2 − v1 v2 − v3 18 − 6 18 − 9
+
=
+
=6A
8
2
8
2
This value of current source current agrees with the value calculated using mesh equations, as it
must.
Figure 8 The circuit from Figure 6 after labeling the nodes. (The encircled numbers are node
numbers.)
Example 3:
Consider the circuit shown in Figure 9. Find the value of the current source current, Ia.
Figure 9 The circuit considered in Example 3.
Solution: Figure 10 shows the circuit from Figure 9 after replacing the ammeter by an equivalent
short circuit and labeling the current measured by the ammeter. This circuit can be analyzed
using mesh equations or using node equations. We will do both.
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First, consider analyzing the circuit in Figure 10 using mesh equations. Figure 11 shows
the circuit after labeling the mesh currents, i1 and i2.
Figure 10 The circuit from Figure 9 after replacing the ammeter by a short circuit.
Figure 11 The circuit from Figure 10 after labeling the meshes.
The mesh current i2 flows through the short circuit that replaced the ammeter. The reference
direction of i2 does not agree with the reference direction of the current measured by the
ammeter, so
i2 = −0.75 A
The mesh current i1 flows through the current source. The reference direction of i2 does not agree
with the reference direction of the current source current, so
i1 = − I a
Apply KVL to the right mesh to get
2 + 6 i2 + 2 ( i2 − i1 ) = 0 ⇒ 2 + 6 ( −0.75 ) + 2 ( −0.75 − ( − I a ) ) = 0 ⇒ I a = 2 A
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Next, consider analyzing the circuit in Figure 10 using node equations. Figure 12 shows
the circuit after selecting a reference node and numbering the other nodes. Let v1 and v2 denote
the node voltages at node 1 and node 2, respectively.
The current measured by the ammeter is equal to the current in the 6 Ω resistor. The
voltage across the 6 Ω resistor can be determined using Ohm’s Law. The voltage across the 6 Ω
resistor can also be expressed in terms of the node voltages. Doing both gives
6 ( 0.75 ) = 0 − v2
⇒ v2 = −4.5 V
The voltage of the voltage source can be expressed in terms of the node voltages as
2 = v1 − v2
⇒ 2 = v1 − ( −4.5 ) ⇒ v1 = −2.5 V
Apply KCL to the supernode to get
Ia +
v1
−2.5
= 0.75 ⇒ I a +
= 0.75 ⇒ I a = 2 A
2
2
As expected, this value of current source current agrees with the value calculated using mesh
equations.
Figure 12 The circuit from Figure 10 after labeling the nodes. (The encircled numbers are node
numbers.)
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Example 4:
Consider the circuit shown in Figure 13. Find the value of the current source current, Ia.
Figure 13 The circuit considered in Example 4.
Solution: Figure 14 shows the circuit from Figure 13 after replacing the voltmeter by an
equivalent open circuit and labeling the voltage measured by the voltmeter. This circuit can be
analyzed using mesh equations or using node equations. We will do both.
First, consider analyzing the circuit in Figure 14 using mesh equations. Figure 15 shows
the circuit after labeling the mesh currents, i1 and i2.
Figure 14 The circuit from Figure 13 after replacing the voltmeter by a open circuit.
Figure 15 The circuit from Figure 14 after labeling the meshes.
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The mesh current i2 is equal to the current in the 4 Ω resistor. The voltmeter measured to voltage
across that 4 Ω resistor to be 9 V. Using Ohm’s Law, the mesh current is given by
i2 =
9
= 2.25 A
4
The mesh current i1 flows through the current source. The reference direction of i2 agrees with
the reference direction of the current source current, so
I a = i1
Apply KVL to the right mesh to get
6 + 9 + 20 i2 − 16 ( i1 − i2 ) = 0 ⇒ 15 + 20 ( 2.25 ) − 16 ( I a − 2.25 ) = 0 ⇒ I a = 6 A
Next, consider analyzing the circuit in Figure 14 using node equations. Figure 16 shows
the circuit after selecting a reference node and numbering the other nodes. Let v1, v2 and v3
denote the node voltages at nodes 1, 2 and 3, respectively.
Apply KCL at node 3 to get
v2 − v3 v3
=
(3)
4
20
The voltage measured by the voltmeter is expressed in terms of the node voltages as
v2 − v3 = 9 V
(4)
Substituting this expression into Equation 3 gives
9 v3
=
⇒ v3 = 45 V
4 20
Substituting this expression into Equation 4 gives
v2 = 54 V
The voltage of the voltage source can be expressed in terms of the node voltages as
6 = v1 − v2
⇒ 6 = v1 − 54 ⇒ v1 = 60 V
Apply KCL to the supernode to get
Ia =
v1 v2 − v3 60 9
+
=
+ =6A
16
4
16 4
As expected, this value of current source current agrees with the value calculated using mesh
equations.
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Figure 16 The circuit from Figure 14 after labeling the nodes. (The encircled numbers are node
numbers.
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