SME
Circuit Analysis I
Node and mesh analysis
Use this for “Vₛ = 10 V, Iₛ = 1 mA, R₁ = 1 kΩ and R₂ = R₃ = R₄ = 2 kΩ: find the node voltages V₁ and V₂ by nodal analysis.”
Diagram
KCL at V₁: 4 = 3 + 1 mA (in = out)
KCL at V₂: 1 + 1 = 2 mA (in = out)
This is an example. Type your own values to start a new problem.
V₁ (V)
−
Vₛ (V)
R₁ (kΩ)
+
V₁ (V)
R₂ (kΩ)
+
V₁ (V)
−
V₂ (V)
R₃ (kΩ)
=
0
V₂ (V)
−
V₁ (V)
R₃ (kΩ)
+
V₂ (V)
R₄ (kΩ)
=
Iₛ (mA)
Each line is KCL at one node. Gathering the V₁ and V₂ terms gives the two-by-two system that the Formulas solve by Cramer’s rule: Δ first, then V₁ and V₂.
Δ
Determinant
worked out (not typed) · mS²
Formulas
Δ
=
(
1
R₁
+
1
R₂
+
1
R₃
)
×
(
1
R₃
+
1
R₄
)
−
(
1
R₃
)
2
1.75
=
(
1
1
+
1
2
+
1
2
)
×
(
1
2
+
1
2
)
−
(
1
2
)
2
V₁
=
(Vₛ
÷
R₁
×
(1
÷
R₃
+
1
÷
R₄
)
+
Iₛ
÷
R₃
)
÷
Δ
6 = (10 ÷ 1 × (1 ÷ 2 + 1 ÷ 2) + 1 ÷ 2) ÷ 1.75
V₂
=
((1
÷
R₁
+
1
÷
R₂
+
1
÷
R₃
)
×
Iₛ
+
Vₛ
÷
(
R₁
×
R₃
))
÷
Δ
4 = ((1 ÷ 1 + 1 ÷ 2 + 1 ÷ 2) × 1 + 10 ÷ (1 × 2)) ÷ 1.75
Assumptions
•
Kirchhoff’s current law at each node: the currents leaving node 1 add to 0, and the currents leaving node 2 add to Iₛ.
•
Ground is the bottom wire (0 V); each conductance is G = 1 ÷ R, in mS with R in kΩ, so Δ is in mS².
•
The system [G₁ + G₂ + G₃, −G₃; −G₃, G₃ + G₄] × [V₁; V₂] = [VₛG₁; Iₛ] is solved by Cramer’s rule.
Step-by-step
WE KNOW
Vₛ = 10 V, Iₛ = 1 mA, R₁ = 1 kΩ, R₂ = 2 kΩ, R₃ = 2 kΩ, R₄ = 2 kΩ
FIND
determinant (Δ), node voltage 1 (V₁), node voltage 2 (V₂)
Step 1 · Find determinant (Δ)
Use Δ = (1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃) × (1 ÷ R₃ + 1 ÷ R₄) − (1 ÷ R₃)²
The determinant of the conductance matrix: the main diagonal’s product minus the product of the two −G₃ entries.
Δ
=
(
1
1
+
1
2
+
1
2
)
×
(
1
2
+
1
2
)
−
(
1
2
)
2
G₁ + G₂ + G₃ = 2 mS and G₃ + G₄ = 1 mS
Δ
=
2
×
1
−
0.5
2
=
1.75
Δ
=
1.75
mS²
Step 2 · Find node voltage 1 (V₁)
Use V₁ = (Vₛ ÷ R₁ × (1 ÷ R₃ + 1 ÷ R₄) + Iₛ ÷ R₃) ÷ Δ
Cramer’s
rule:
put
the
right
side
[VₛG₁;
Iₛ]
in
the
first
column,
take
that
determinant
Δ₁,
then
divide
by
Δ
.
V₁
=
(10
÷
1
×
(1
÷
2
+
1
÷
2)
+
1
÷
2)
÷
1.75
Δ₁ = 10 × 1 − (−0.5) × 1 = 10.5
V₁
=
10.5
1.75
V₁
=
6
V
Step 3 · Find node voltage 2 (V₂)
Use V₂ = ((1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃) × Iₛ + Vₛ ÷ (R₁ × R₃)) ÷ Δ
Cramer’s
rule:
put
the
right
side
in
the
second
column,
take
that
determinant
Δ₂,
then
divide
by
Δ
.
V₂
=
((1
÷
1
+
1
÷
2
+
1
÷
2)
×
1
+
10
÷
(1
×
2))
÷
1.75
Δ₂ = 2 × 1 − (−0.5) × 10 = 7
V₂
=
7
1.75
V₂
=
4
V
Check
Put the numbers back in. Both sides should match.
1.75
=
(
1
1
+
1
2
+
1
2
)
×
(
1
2
+
1
2
)
−
(
1
2
)
2
✓
6 = (10 ÷ 1 × (1 ÷ 2 + 1 ÷ 2) + 1 ÷ 2) ÷ 1.75 ✓
4 = ((1 ÷ 1 + 1 ÷ 2 + 1 ÷ 2) × 1 + 10 ÷ (1 × 2)) ÷ 1.75 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.