SME
Circuit Analysis I · Problem type
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?”
Diagram
KVL: 8 = 2 + 6 V (rises = drops)
This is an example. Type your own values to start a new problem.
PL (W)
=
(
VTh (V)
)
2
×
RL (Ω)
(
RTh (Ω)
+
RL (Ω)
)
2
Pₘₐₓ (W)
=
(
VTh (V)
)
2
4
×
RTh (Ω)
Formulas
PL
=
VTh
2
×
RL
÷
(RTh + RL)
2
6
=
8
2
×
6
÷
(2 + 6)
2
Pₘₐₓ
=
VTh
2
4 × RTh
8
=
8
2
4 × 2
Assumptions
•
The load sees the circuit as its Thévenin equivalent: VTh in series with RTh.
•
The load current is VTh ÷ (RTh + RL), and the load takes PL = I²RL.
•
The load takes the most power when RL = RTh; RTh then wastes just as much.
Step-by-step
WE KNOW
VTh = 8 V, RTh = 2 Ω, RL = 6 Ω
FIND
load power (PL), maximum load power (Pₘₐₓ)
Step 1 · Find load power (PL)
Use PL = VTh² × RL ÷ (RTh + RL)²
The
loop
current
VTh
RTh + RL
,
squared,
times
RL.
PL
=
8
2
×
6
÷
(2 + 6)
2
PL
=
8
2
×
6
8
2
PL
=
64
×
6
64
PL
=
384
64
PL = 6 W
Step 2 · Find maximum load power (Pₘₐₓ)
Use Pₘₐₓ = VTh² ÷ (4 × RTh)
At
RL
=
RTh
the
load
gets
half
of
VTh,
so
Pₘₐₓ
=
(
VTh
2
)
2
÷
RTh.
Pₘₐₓ
=
8
2
4 × 2
Pₘₐₓ
=
8
2
8
Pₘₐₓ
=
64
8
Pₘₐₓ = 8 W
Check
Put the numbers back in. Both sides should match.
6
=
8
2
×
6
÷
(2 + 6)
2
✓
8
=
8
2
4 × 2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Thévenin and Norton equivalents
Main lesson
Norton equivalent
Use this for “Turn an 8 V, 2 Ω Thévenin equivalent into its Norton form.”
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.”