SME
Circuit Analysis I · Problem type
Superposition of two sources
Use this for “A 12 V source through 3 Ω and a 3 A current source feed one node, with 6 Ω to ground. Find the node voltage by superposition.”
Diagram
KCL at V: 3 = 0.667 + 2.33 A (in = out)
V = V′ + V″ = 8 V + 6 V = 14 V
This is an example. Type your own values to start a new problem.
V′ (V)
=
Vₛ (V)
×
R₂ (Ω)
R₁ (Ω)
+
R₂ (Ω)
V″ (V)
=
Iₛ (A)
×
R₁ (Ω)
×
R₂ (Ω)
R₁ (Ω)
+
R₂ (Ω)
V (V)
=
V′ (V)
+
V″ (V)
Formulas
V
′
=
Vₛ
×
R₂
R₁
+
R₂
8
=
12
×
6
3 + 6
V
″
=
Iₛ
×
R₁
×
R₂
R₁
+
R₂
6
=
3
×
3
×
6
3 + 6
V
=
V
′
+
V
″
14 = 8 + 6
Assumptions
•
The circuit is linear, so the node voltage with both sources is the sum of each source acting alone.
•
A voltage source is turned off by a short, a current source by an open.
•
Iₛ flows into the node; a negative Iₛ flows out of it.
Step-by-step
WE KNOW
Vₛ = 12 V, Iₛ = 3 A, R₁ = 3 Ω, R₂ = 6 Ω
FIND
voltage from Vₛ alone (V′), voltage from Iₛ alone (V″), node voltage (V)
Step 1 · Find voltage from Vₛ alone (V′)
Use V′ = Vₛ × R₂ ÷ (R₁ + R₂)
Open
Iₛ:
R₁
and
R₂
are
in
series
and
divide
Vₛ.
V
′
=
12
×
6
3 + 6
V
′
=
12
×
6
9
V
′
=
72
9
V
′
=
8
V
Step 2 · Find voltage from Iₛ alone (V″)
Use V″ = Iₛ × R₁ × R₂ ÷ (R₁ + R₂)
Short
Vₛ:
R₁
and
R₂
both
run
from
the
node
to
ground,
so
Iₛ
flows
into
them
in
parallel.
V
″
=
3
×
3
×
6
3 + 6
V
″
=
3
×
3
×
6
9
V
″
=
9
×
6
9
V
″
=
54
9
V
″
=
6
V
Step 3 · Find node voltage (V)
Use V = V′ + V″
Superposition: add the two one-source answers.
V
=
8
+
6
V
=
14
V
Check
Put the numbers back in. Both sides should match.
8
=
12
×
6
3 + 6
✓
6
=
3
×
3
×
6
3 + 6
✓
14 = 8 + 6 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.