SME
Circuit Analysis I · Problem type
Series-parallel reduction
Use this for “R₁ = 4 Ω is in series with 6 Ω ∥ 12 Ω across 24 V. Find the total resistance and every current.”
Diagram
KCL at the junction: 3 = 2 + 1 A (in = out)
KVL round loop 1: 24 = 12 + 12 V (rises = drops)
KVL round loop 2: 12 = 12 V (rises = drops)
This is an example. Type your own values to start a new problem.
Rₚ (Ω)
=
R₂ (Ω)
×
R₃ (Ω)
R₂ (Ω)
+
R₃ (Ω)
Rₜ (Ω)
=
R₁ (Ω)
+
Rₚ (Ω)
V (V)
=
I (A)
×
Rₜ (Ω)
V₁
Voltage across R₁
calculated
I₂
Current in R₂
calculated
I₃
Current in R₃
calculated
Formulas
Rₚ
=
R₂
×
R₃
R₂
+
R₃
4
=
6
×
12
6 + 12
Rₜ
=
R₁
+
Rₚ
8 = 4 + 4
V
=
I
×
Rₜ
24 = 3 × 8
V₁
=
I
×
R₁
12 = 3 × 4
I₂
=
I
×
Rₚ
R₂
2
=
3
×
4
6
I₃
=
I
×
Rₚ
R₃
1
=
3
×
4
12
Assumptions
•
R₂ and R₃ are in parallel, so they share one voltage; that pair is in series with R₁.
•
Reduce from the inside out: the pair to Rₚ, then Rₚ plus R₁ gives the total Rₜ.
•
Kirchhoff’s current law: the source current I splits at the junction into I₂ and I₃.
Step-by-step
WE KNOW
V = 24 V, R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω
FIND
parallel pair R₂ ∥ R₃ (Rₚ), total resistance (Rₜ), source current (I), voltage across R₁ (V₁), current in R₂ (I₂), current in R₃ (I₃)
Step 1 · Find parallel pair R₂ ∥ R₃ (Rₚ)
Use Rₚ = R₂ × R₃ ÷ (R₂ + R₃)
Product
over
sum:
R₂
and
R₃
in
parallel
reduce
to
one
resistance.
Rₚ
=
6
×
12
6 + 12
Rₚ
=
6
×
12
18
Rₚ
=
72
18
Rₚ = 4 Ω
Step 2 · Find total resistance (Rₜ)
Use Rₜ = R₁ + Rₚ
R₁
is
in
series
with
the
pair,
so
the
resistances
add.
Rₜ
=
4
+
4
Rₜ
=
8
Ω
Step 3 · Find source current (I)
Use V = I × Rₜ
Ohm’s law for the reduced circuit: source voltage over total resistance.
I
=
V
Rₜ
I
=
24
8
I
=
3
A
Step 4 · Find voltage across R₁ (V₁)
Use V₁ = I × R₁
All
the
source
current
passes
through
R₁
;
Ohm’s
law
gives
its
drop.
V₁
=
3
×
4
V₁
=
12
V
Step 5 · Find current in R₂ (I₂)
Use I₂ = I × Rₚ ÷ R₂
The
pair
has
I
×
Rₚ
across
it
(the
rest
of
V
after
V₁
);
Ohm’s
law
for
R₂
.
I₂
=
3
×
4
6
I₂
=
12
6
I₂
=
2
A
Step 6 · Find current in R₃ (I₃)
Use I₃ = I × Rₚ ÷ R₃
R₃
has
the
same
voltage
as
R₂
;
the
two
branch
currents
add
back
to
I
(KCL).
I₃
=
3
×
4
12
I₃
=
12
12
I₃
=
1
A
Check
Put the numbers back in. Both sides should match.
4
=
6
×
12
6 + 12
✓
8 = 4 + 4 ✓
24 = 3 × 8 ✓
12 = 3 × 4 ✓
2
=
3
×
4
6
✓
1
=
3
×
4
12
✓
Topic check
See what you know: 5 questions on this topic
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