SME
Circuit Analysis I
RC and RL transients
Use this for “A 100 μF capacitor charges through 10 kΩ from 10 V. Find vC after 1 s.”
Diagram
The switch closes at t = 0; the capacitor’s voltage can’t jump, so it starts at its old value.
This is an example. Type your own values to start a new problem.
τ (s)
=
R (kΩ)
×
C (μF)
vC (V)
=
Vₛ (V)
×
(1
−
e
−
s
÷
s
)
i
Current
calculated
Formulas
τ
=
R
×
C
1,000 = 10 × 100
vC
=
Vₛ
×
(1
−
e
−
t
÷
τ
)
6.3212
=
10
×
(1
−
e
−1,000 ÷ 1,000
)
i
=
Vₛ − vC
R
0.3679
=
10 − 6.3212
10
Formulas are worked in ms. The step-by-step shows the conversions.
Assumptions
•
The capacitor starts empty and the switch closes at t = 0; the source and the wires are ideal.
•
The capacitor charges as vC = Vₛ(1 − e−t/τ) with τ = RC; after 5τ it is within 1% of Vₛ.
•
kΩ × μF gives ms, and V ÷ kΩ gives mA.
Step-by-step
WE KNOW
Vₛ = 10 V, R = 10 kΩ, C = 100 μF, t = 1 s
FIND
time constant (τ), capacitor voltage (vC), current (i)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to ms).
t = 1 s = 1,000 ms (1 s = 1000 ms)
Step 2 · Find time constant (τ)
Use τ = R × C
The time constant is RC: kΩ × μF gives ms.
τ
=
10
×
100
τ
=
1,000
ms
Step 3 · Find capacitor voltage (vC)
Use vC = Vₛ × (1 − e−t ÷ τ)
The
share
of
the
climb
to
Vₛ
made
so
far:
63.2%
after
one
τ
,
86.5%
after
2
τ
.
vC
=
10
×
(1
−
e
−1,000 ÷ 1,000
)
vC = 10 × (1 − 0.3679)
vC = 10 × 0.6321
vC = 6.3212 V
Step 4 · Find current (i)
Use i = (Vₛ − vC) ÷ R
Kirchhoff’s
voltage
law:
the
resistor
drops
what
the
capacitor
has
not
taken
yet;
V
kΩ
gives
mA.
i
=
10 − 6.3212
10
i
=
3.6788
10
i
=
0.3679
mA
Step 5 · Convert the answers
Change each answer back to the unit you chose.
τ = 1,000 ms = 1 s (1 s = 1000 ms)
Check
Put the numbers back in. Both sides should match.
1,000 = 10 × 100 ✓
6.3212
=
10
×
(1
−
e
−1,000 ÷ 1,000
)
✓
0.3679
=
10 − 6.3212
10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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