SME
University Physics II: Electricity & Magnetism · Problem type
A battery’s internal resistance
Use this for “A 9 V battery with 0.3 Ω inside lights a 2.7 Ω bulb. Find the current, the terminal voltage and the bulb’s power.”
Diagram
KVL: 12 = 1 + 11 V (rises = drops)
This is an example. Type your own values to start a new problem.
I (
)
=
ε (V)
R (
)
+
r (
)
V
Terminal voltage
calculated
P
Power in the load
calculated
Formulas
I
=
ε
R
+
r
2
=
12
5.5 + 0.5
V
=
I
×
R
11 = 2 × 5.5
V
=
ε
−
I
×
r
11 = 12 − 2 × 0.5
P
=
I
2
×
R
22
=
2
2
×
5.5
Assumptions
•
A real battery is an ideal emf ε in series with a small internal resistance r; the wires are ideal.
•
One loop, so one current: I = ε ÷ (R + r), the emf over the load and r together.
•
The terminals sit across the load, so the terminal voltage is V = IR; it is also ε − Ir, the emf less the drop inside the battery.
•
The load takes P = I²R, the most when R = r; the battery then wastes as much as it delivers.
Step-by-step
WE KNOW
ε = 12 V, r = 0.5 Ω, R = 5.5 Ω
FIND
current (I), terminal voltage (V), power in the load (P)
Step 1 · Find current (I)
Use I = ε ÷ (R + r)
One loop: the emf drives the current through the load and the internal resistance in series.
I
=
12
5.5 + 0.5
I
=
12
6
I
=
2
A
Step 2 · Find terminal voltage (V)
Use V = I × R
The terminals sit across the load, so the terminal voltage is the load’s drop IR.
V
=
2
×
5.5
V
=
11
V
Step 3 · Find power in the load (P)
Use P = I² × R
The power the load takes.
P
=
2
2
×
5.5
P
=
4
×
5.5
P
=
22
W
Check
Put the numbers back in. Both sides should match.
2
=
12
5.5 + 0.5
✓
11 = 2 × 5.5 ✓
11 = 12 − 2 × 0.5 ✓
22
=
2
2
×
5.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.