SME
Circuit Analysis I · Problem type
Discharging a capacitor
Use this for “A 50 μF capacitor charged to 9 V discharges through 2 kΩ. How long until it reads 3 V?”
Chart
v(t) = 9exp(−t/τ), τ = 0.1 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 0.1099 s: v = 3 V.
This is an example. Type your own values to start a new problem.
τ (s)
=
R (kΩ)
×
C (μF)
v (V)
=
V₀ (V)
×
e
−
s
÷
s
Formulas
τ
=
R
×
C
100 = 2 × 50
v
=
V₀
×
e
−
t
÷
τ
3
=
9
×
e
−109.8612 ÷ 100
Formulas are worked in ms. The step-by-step shows the conversions.
Assumptions
•
The capacitor holds V₀ when the switch closes at t = 0, and the resistor is its only path.
•
The capacitor discharges as v = V₀e−t/τ with τ = RC: every τ it loses 63.2% of what is left.
•
kΩ × μF gives ms.
Step-by-step
WE KNOW
V₀ = 9 V, R = 2 kΩ, C = 50 μF, v = 3 V
FIND
time constant (τ), time since the switch closed (t)
Step 1 · Find time constant (τ)
Use τ = R × C
The time constant is RC: kΩ × μF gives ms.
τ
=
2
×
50
τ
=
100
ms
Step 2 · Find time since the switch closed (t)
Use v = V₀ × e−t ÷ τ
Take
ln
of
V₀
v
=
e
t
τ
,
then
multiply
by
τ
.
t
=
τ
×
ln(
V₀
v
)
t
=
100
×
ln(
9
3
)
t
=
100
×
ln(3)
t
=
100
×
1.0986
t
=
109.8612
ms
Step 3 · Convert the answers
Change each answer back to the unit you chose.
τ = 100 ms = 0.1 s (1 s = 1000 ms)
t = 109.8612 ms = 0.1099 s (1 s = 1000 ms)
Check
Put the numbers back in. Both sides should match.
100 = 2 × 50 ✓
3
=
9
×
e
−109.8612 ÷ 100
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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