SME
Circuit Analysis I · Problem type
Supernode
Use this for “A 6 V source joins node 1 to node 2; 6 A enters node 1, and R₁ = 2 Ω and R₂ = 4 Ω go to ground. Find V₁ and V₂.”
Diagram
KCL round the supernode: 6 = 5 + 1 A
V₁ − V₂ = 10 − 4 = 6 V, the source between them
This is an example. Type your own values to start a new problem.
V₁ (V)
−
V₂ (V)
=
Vₓ (V)
Iₛ (A)
=
V₁ (V)
R₁ (Ω)
+
V₂ (V)
R₂ (Ω)
Formulas
V₁
=
Iₛ ×
R₁
×
R₂
+ Vₓ ×
R₁
R₁
+
R₂
10
=
6 × 2 × 4 + 6 × 2
2 + 4
V₂
=
V₁
−
Vₓ
4 = 10 − 6
Iₛ
=
V₁
R₁
+
V₂
R₂
6
=
10
2
+
4
4
Assumptions
•
The source between the nodes fixes V₁ − V₂ = Vₓ (+ at node 1); the two nodes and the source form one supernode.
•
KCL round the supernode: the current Iₛ coming in equals V₁ ÷ R₁ + V₂ ÷ R₂ going out to ground.
•
Ground is the bottom wire (0 V); no current equation is written for the source’s own current.
Step-by-step
WE KNOW
Vₓ = 6 V, Iₛ = 6 A, R₁ = 2 Ω, R₂ = 4 Ω
FIND
node voltage 1 (V₁), node voltage 2 (V₂)
Step 1 · Find node voltage 1 (V₁)
Use V₁ = (Iₛ × R₁ × R₂ + Vₓ × R₁) ÷ (R₁ + R₂)
Put
V₂
=
V₁
−
Vₓ
into
V₁
R₁
+
V₂
R₂
=
Iₛ,
multiply
through
by
R₁
R₂
and
collect
V₁
.
V₁
=
6 × 2 × 4 + 6 × 2
2 + 4
V₁
=
12 × 4 + 12
6
V₁
=
48 + 12
6
V₁
=
60
6
V₁
=
10
V
Step 2 · Find node voltage 2 (V₂)
Use V₂ = V₁ − Vₓ
The source between the nodes sets their difference, so node 2 sits Vₓ below node 1.
V₂
=
10
−
6
V₂
=
4
V
Check
Put the numbers back in. Both sides should match.
10
=
6 × 2 × 4 + 6 × 2
2 + 4
✓
4 = 10 − 6 ✓
6
=
10
2
+
4
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.