SME
Circuit Analysis I
Thévenin and Norton equivalents
Use this for “A 12 V source feeds a 3 Ω series resistor and a 6 Ω shunt. Find the Thévenin equivalent seen by a 6 Ω load, and the load current.”
Diagram
KCL at the junction: 2 = 1 + 1 A (in = out)
VTh
=
Vₛ
R₂
R₁ + R₂
=
8
V
RTh
=
R₁
R₂
R₁ + R₂
,
Vₛ
shorted,
=
2
Ω
IL
=
VTh
RTh + RL
=
8
V
2 Ω + 6 Ω
IL = 1 A
This is an example. Type your own values to start a new problem.
VTh (V)
=
Vₛ (V)
×
R₂ (Ω)
R₁ (Ω)
+
R₂ (Ω)
RTh (Ω)
=
R₁ (Ω)
×
R₂ (Ω)
R₁ (Ω)
+
R₂ (Ω)
IL (A)
=
VTh (V)
RTh (Ω)
+
RL (Ω)
VL
Load voltage
calculated
Formulas
VTh
=
Vₛ
×
R₂
R₁
+
R₂
8
=
12
×
6
3 + 6
RTh
=
R₁
×
R₂
R₁
+
R₂
2
=
3
×
6
3 + 6
IL
=
VTh
RTh + RL
1
=
8
2 + 6
VL = IL × RL
6 = 1 × 6
Assumptions
•
VTh is the open-circuit voltage at the load terminals a–b; RTh is the resistance seen there with Vₛ zeroed.
•
A voltage source is zeroed by replacing it with a short, which puts R₁ and R₂ in parallel.
•
The circuit is linear, so the equivalent gives every load the same current as the full circuit.
Step-by-step
WE KNOW
Vₛ = 12 V, R₁ = 3 Ω, R₂ = 6 Ω, RL = 6 Ω
FIND
Thévenin voltage (VTh), Thévenin resistance (RTh), load current (IL), load voltage (VL)
Step 1 · Find Thévenin voltage (VTh)
Use VTh = Vₛ × R₂ ÷ (R₁ + R₂)
With
the
load
removed
no
current
leaves
at
a–b,
so
R₁
and
R₂
divide
the
source
voltage.
VTh
=
12
×
6
3 + 6
VTh
=
12
×
6
9
VTh
=
72
9
VTh = 8 V
Step 2 · Find Thévenin resistance (RTh)
Use RTh = R₁ × R₂ ÷ (R₁ + R₂)
Short
the
source:
looking
in
from
a–b,
R₁
and
R₂
are
in
parallel
(product
over
sum).
RTh
=
3
×
6
3 + 6
RTh
=
3
×
6
9
RTh
=
18
9
RTh = 2 Ω
Step 3 · Find load current (IL)
Use IL = VTh ÷ (RTh + RL)
The equivalent is one loop: VTh drives RTh and the load in series.
IL
=
8
2 + 6
IL
=
8
8
IL = 1 A
Step 4 · Find load voltage (VL)
Use VL = IL × RL
Ohm’s law for the load.
VL = 1 × 6
VL = 6 V
Check
Put the numbers back in. Both sides should match.
8
=
12
×
6
3 + 6
✓
2
=
3
×
6
3 + 6
✓
1
=
8
2 + 6
✓
6 = 1 × 6 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Norton equivalent
Use this for “Turn an 8 V, 2 Ω Thévenin equivalent into its Norton form.”
Maximum power transfer
Use this for “An 8 V, 2 Ω Thévenin source drives a 6 Ω load. What power does the load take, and what is the most any load can take?”
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.”