SME
Circuit Analysis I · Problem type
Mesh analysis
Use this for “Two meshes share R₂ = 4 Ω, with R₁ = R₃ = 2 Ω, Vₐ = 8 V and Vb = 2 V. Find I₁, I₂ and the current in R₂.”
Diagram
KVL round loop 1: 8 = 4 + 4 V (rises = drops)
KVL round loop 2: 4 = 2 + 2 V (rises = drops)
This is an example. Type your own values to start a new problem.
(
R₁ (Ω)
+
R₂ (Ω)
)
×
I₁ (A)
−
R₂ (Ω)
×
I₂ (A)
=
Vₐ (V)
−
R₂ (Ω)
×
I₁ (A)
+
(
R₂ (Ω)
+
R₃ (Ω)
)
×
I₂ (A)
=
−
Vb (V)
Each line is KVL round one mesh. The Formulas solve the pair by Cramer’s rule: Δ first, then I₁ and I₂.
Δ
Determinant
worked out (not typed) · Ω²
IR2
Current in R₂
calculated
Formulas
Δ
=
(
R₁
+
R₂
)
×
(
R₂
+
R₃
)
−
R₂
2
20
=
(2
+
4)
×
(4
+
2)
−
4
2
I₁
=
Vₐ × (
R₂
+
R₃
) −
R₂
× Vb
Δ
2
=
8 × (4 + 2) − 4 × 2
20
I₂
=
R₂
× Vₐ − (
R₁
+
R₂
) × Vb
Δ
1
=
4 × 8 − (2 + 4) × 2
20
IR2
=
I₁
−
I₂
1 = 2 − 1
Assumptions
•
Both mesh currents are taken clockwise; Vb’s + terminal meets I₂ first.
•
KVL round each mesh: (R₁ + R₂)I₁ − R₂I₂ = Vₐ and −R₂I₁ + (R₂ + R₃)I₂ = −Vb, solved by Cramer’s rule (Δ in Ω²).
•
The shared resistor R₂ carries I₁ − I₂ downward; a negative value means it flows up.
Step-by-step
WE KNOW
Vₐ = 8 V, Vb = 2 V, R₁ = 2 Ω, R₂ = 4 Ω, R₃ = 2 Ω
FIND
determinant (Δ), mesh current 1 (I₁), mesh current 2 (I₂), current in R₂ (IR2)
Step 1 · Find determinant (Δ)
Use Δ = (R₁ + R₂) × (R₂ + R₃) − R₂²
The
determinant
of
the
resistance
matrix
[
R₁
+
R₂
,
−
R₂
;
−
R₂
,
R₂
+
R₃
]:
the
diagonal’s
product
minus
R₂
2
.
Δ
=
(2
+
4)
×
(4
+
2)
−
4
2
Δ
=
6
×
6
−
4
2
Δ
=
6
×
6
−
16
Δ
=
36
−
16
Δ
=
20
Ω²
Step 2 · Find mesh current 1 (I₁)
Use I₁ = (Vₐ × (R₂ + R₃) − R₂ × Vb) ÷ Δ
Cramer’s
rule:
put
the
right
side
[Vₐ;
−Vb]
in
the
first
column,
take
that
determinant
Δ₁,
then
divide
by
Δ
.
I₁
=
8 × (4 + 2) − 4 × 2
20
Δ₁ = 8 × 6 − (−4) × (−2) = 40
I₁
=
40
20
I₁
=
2
A
Step 3 · Find mesh current 2 (I₂)
Use I₂ = (R₂ × Vₐ − (R₁ + R₂) × Vb) ÷ Δ
Cramer’s
rule:
put
the
right
side
in
the
second
column,
take
that
determinant
Δ₂,
then
divide
by
the
same
Δ
.
I₂
=
4 × 8 − (2 + 4) × 2
20
Δ₂ = 6 × (−2) − (−4) × 8 = 20
I₂
=
20
20
I₂
=
1
A
Step 4 · Find current in R₂ (IR2)
Use IR2 = I₁ − I₂
Both
mesh
currents
pass
through
R₂
,
in
opposite
directions,
so
the
branch
carries
their
difference.
IR2 = 2 − 1
IR2 = 1 A
Check
Put the numbers back in. Both sides should match.
20
=
(2
+
4)
×
(4
+
2)
−
4
2
✓
2
=
8 × (4 + 2) − 4 × 2
20
✓
1
=
4 × 8 − (2 + 4) × 2
20
✓
1 = 2 − 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.