SME
Circuit Analysis I · Problem type

Switching from a nonzero start

Use this for “A capacitor voltage starts at 2 V and heads for 10 V with τ = 0.5 ms. Find it at 1 ms.”

Chart

00.511.522.50510v (V)Time t (ms)final 10 V63.2%τ2τ3τ4τ5τv = 8.917 V
v(t) = 10 + (2 − 10)exp(−t/τ), τ = 0.5 ms
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 1 ms: v = 8.917 V.
This is an example. Type your own values to start a new problem.
v (V)
=
vf (V)
+
(
v₀ (V)
−
vf (V)
)
×
e
−
ms
÷
ms

Formulas

v
=
vf
+
(
v₀
−
vf)
×
e
−
t
÷
τ
8.9173
=
10
+
(2
−
10)
×
e
−1 ÷ 0.5

Assumptions

•
One energy store (one C or one L), so every voltage and current is first order.
•
v(t) = vf + (v₀ − vf)e−t/τ, with t counted from the switching; a current works the same way.
•
v₀ is the value just after switching, vf the value long after, and τ is RC or L ÷ R seen from the store.

Step-by-step

WE KNOW
v₀ = 2 V, vf = 10 V, τ = 0.5 ms, t = 1 ms
FIND
value at t (v)
Step 1 · Find value at t (v)
Use v = vf + (v₀ − vf) × e−t ÷ τ
Start
from
the
final
value
and
add
the
gap
still
left:
it
shrinks
to
36.8%
every
τ
.
v
=
10
+
(2
−
10)
×
e
−1 ÷ 0.5
v
=
10
+
(−8)
×
e
-2
v
=
10
+
(−8)
×
0.1353
v
=
10
+
(−1.0824)
v
=
8.9173
V
Check
Put the numbers back in. Both sides should match.
8.9173
=
10
+
(2
−
10)
×
e
−1 ÷ 0.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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