SME
University Physics II: Electricity & Magnetism
Potential and capacitance
Use this for “A 3 μF capacitor is in series with 4 μF and 2 μF in parallel, across 18 V. Find the charge, the energy and each voltage.”
Diagram
KCL at the junction: 24 = 16 + 8 μC (in = out)
KVL round loop 1: 12 = 4 + 8 V (rises = drops)
KVL round loop 2: 8 = 8 V (rises = drops)
This is an example. Type your own values to start a new problem.
Ceq (
)
=
C₁ (
)
(
C₂ (
)
+
C₃ (
)
)
+
+
Q (
)
=
Ceq (
)
×
V (V)
U
Energy stored
calculated · J
V₁
Voltage on C₁
calculated
V₂
Voltage on C₂ and C₃
calculated
Formulas
Ceq
=
C₁
×
C₂
+
C₃
C₁
+
C₂
+
C₃
2
=
6
×
2 + 1
6 + 2 + 1
Q
=
Ceq
×
V
24 = 2 × 12
U
=
1
2
×
Ceq
×
10
-6
×
V
2
1.44 ×
10
-4
=
1
2
×
2 ×
10
-6
×
12
2
V₁
=
Q
C₁
4
=
24
6
V₂
=
Q
C₂
+
C₃
8
=
24
2 + 1
Assumptions
•
In parallel the capacitors share one voltage, so their capacitances add: C₂ ∥ C₃ = C₂ + C₃.
•
In series each capacitor holds the same charge Q and 1 ÷ C adds, so C₁ with the pair gives Ceq = C₁(C₂ + C₃) ÷ (C₁ + C₂ + C₃).
•
The capacitors start uncharged and end fully charged; μF × V gives μC.
Step-by-step
WE KNOW
C₁ = 6 μF, C₂ = 2 μF, C₃ = 1 μF, V = 12 V
FIND
equivalent capacitance (Ceq), charge on C₁ (Q), energy stored (U), voltage on C₁ (V₁), voltage on C₂ and C₃ (V₂)
Step 1 · Find equivalent capacitance (Ceq)
Use Ceq = C₁ × (C₂ + C₃) ÷ (C₁ + C₂ + C₃)
C₂
and
C₃
add
in
parallel;
that
pair
in
series
with
C₁
is
product
over
sum.
Ceq
=
6
×
2 + 1
6 + 2 + 1
Ceq
=
6
×
3
8 + 1
Ceq
=
6
×
3
9
Ceq
=
18
9
Ceq = 2 μF
Step 2 · Find charge on C₁ (Q)
Use Q = Ceq × V
The
charge
the
source
moves
onto
the
network,
the
same
on
C₁
as
on
the
pair.
Q
=
2
×
12
Q
=
24
μC
Step 3 · Find energy stored (U)
Use U = ½ × Ceq × 10⁻⁶ × V²
The
energy
in
the
whole
network
is
the
energy
in
Ceq:
half
C
in
farads
times
V
2
.
U
=
1
2
×
2 ×
10
-6
×
12
2
U
=
1
2
×
(2
×
10
-6
)
×
144
U
=
10
-6
×
144
U
=
1.44 ×
10
-4
J
Step 4 · Find voltage on C₁ (V₁)
Use V₁ = Q ÷ C₁
C₁
is
in
series
with
the
source,
so
it
holds
the
full
charge
Q
.
V₁
=
24
6
V₁
=
4
V
Step 5 · Find voltage on C₂ and C₃ (V₂)
Use V₂ = Q ÷ (C₂ + C₃)
The
pair
holds
the
same
Q
as
C₁
,
shared
between
C₂
and
C₃
at
one
voltage.
V₂
=
24
2 + 1
V₂
=
24
3
V₂
=
8
V
Check
Put the numbers back in. Both sides should match.
2
=
6
×
2 + 1
6 + 2 + 1
✓
24 = 2 × 12 ✓
1.44 ×
10
-4
=
1
2
×
2 ×
10
-6
×
12
2
✓
4
=
24
6
✓
8
=
24
2 + 1
✓
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The capacitance of a coaxial cable
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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