SME
Grade 11 Science · Problem type
Resistors in series
Use this for “A 12 V battery drives 10 Ω and 20 Ω in series. What does the ammeter read, and what is the voltage across each?”
Diagram
V = V₁ + V₂ = 4 V + 8 V = 12 V
I
=
V
R₁ + R₂
=
12
V
10 Ω + 20 Ω
I = 0.4 A
Drag the source or a resistor up or down to change it.
Total resistance: 30 Ω
This is an example. Type your own values to start a new problem.
V
Battery voltage
example
R₁
First resistor
example
R₂
Second resistor
example
R
Total resistance
calculated
I
Current
calculated
V₁
Voltage across R₁
calculated
V₂
Voltage across R₂
calculated
Formulas
R
=
R₁
+
R₂
30 = 10 + 20
I
=
V
R
0.4
=
12
30
V₁
=
I
×
R₁
4 = 0.4 × 10
V₂
=
I
×
R₂
8 = 0.4 × 20
Assumptions
•
In series there is one path, so the same current I flows through each resistor.
•
Series resistances add: R = R₁ + R₂.
•
The battery’s voltage is shared in the ratio of the resistances: V₁ + V₂ = V.
Step-by-step
WE KNOW
V = 12 V, R₁ = 10 Ω, R₂ = 20 Ω
FIND
total resistance (R), current (I), voltage across R₁ (V₁), voltage across R₂ (V₂)
Step 1 · Find total resistance (R)
Use R = R₁ + R₂
The current passes through both: the resistances add.
R
=
10
+
20
R
=
30
Ω
Step 2 · Find current (I)
Use I = V/R
Ohm’s law for the whole loop: the battery voltage over the total resistance.
I
=
12
30
I
=
0.4
A
Step 3 · Find voltage across R₁ (V₁)
Use V₁ = I × R₁
Ohm’s
law
for
R₁
alone:
the
current
times
its
resistance.
V₁
=
0.4
×
10
V₁
=
4
V
Step 4 · Find voltage across R₂ (V₂)
Use V₂ = I × R₂
Ohm’s
law
for
R₂
alone:
the
current
times
its
resistance.
V₂
=
0.4
×
20
V₂
=
8
V
Check
Put the numbers back in. Both sides should match.
30 = 10 + 20 ✓
0.4
=
12
30
✓
4 = 0.4 × 10 ✓
8 = 0.4 × 20 ✓
Topic check
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