SME
Circuit Analysis I · Problem type
Current in an RL circuit
Use this for “A 2 H coil and a 4 Ω resistor are switched onto 12 V. Find the current after 1 s.”
Diagram
The switch closes at t = 0; the inductor’s current 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)
=
L (H)
R (Ω)
If (A)
=
Vₛ (V)
R (Ω)
i (A)
=
If (A)
×
(1
−
e
−
s
÷
s
)
Formulas
τ
=
L
R
0.5
=
2
4
If
=
Vₛ
R
3
=
12
4
i
=
If
×
(1
−
e
−
t
÷
τ
)
2.594
=
3
×
(1
−
e
−1 ÷ 0.5
)
Assumptions
•
No current flows before the switch closes at t = 0; the source, coil and wires are ideal.
•
The current climbs as i = If(1 − e−t/τ) with τ = L ÷ R, toward If = Vₛ ÷ R once the coil drops no voltage.
•
mH ÷ Ω gives ms.
Step-by-step
WE KNOW
Vₛ = 12 V, R = 4 Ω, L = 2 H, t = 1 s
FIND
time constant (τ), final current (If), current at t (i)
Step 1 · Find time constant (τ)
Use τ = L ÷ R
The
time
constant
is
L
over
R
:
mH
Ω
gives
ms.
τ
=
2
4
τ
=
0.5
s
Step 2 · Find final current (If)
Use If = Vₛ ÷ R
Once
the
current
is
steady
the
coil
drops
no
voltage,
so
all
of
Vₛ
is
across
R
.
If
=
12
4
If = 3 A
Step 3 · Find current at t (i)
Use i = If × (1 − e−t ÷ τ)
The
share
of
the
final
current
reached
so
far:
63.2%
after
one
τ
,
86.5%
after
2
τ
.
i
=
3
×
(1
−
e
−1 ÷ 0.5
)
i
=
3
×
(1
−
e
-2
)
i
=
3
×
(1
−
0.1353)
i
=
3
×
0.8647
i
=
2.594
A
Check
Put the numbers back in. Both sides should match.
0.5
=
2
4
✓
3
=
12
4
✓
2.594
=
3
×
(1
−
e
−1 ÷ 0.5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
RC and RL transients
Main lesson
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?”
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.”
Damping of a series RLC circuit
Use this for “A series RLC has 10 Ω, 1 mH and 10 μF. Is it over-, under- or critically damped?”