SME
University Physics II: Electricity & Magnetism · Problem type
Current rising in an RL circuit
Use this for “A 24 V battery is switched into a 4 Ω resistor and a 2 H coil. What is the current after 0.25 s, and how much energy does the coil hold at the end?”
Chart
I(t) = 2(1 − exp(−t/τ)), τ = 0.5 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 0.5 s: I = 1.264 A.
This is an example. Type your own values to start a new problem.
τ (
)
=
L (
)
R (
)
I (
)
=
ε (V)
R (
)
×
(1
−
e
−
s
÷
s
)
If
Final current ε ÷ R
calculated
U
Energy stored at the end
calculated · J
Formulas
τ
=
L
R
0.5
=
3
6
If
=
ε
R
2
=
12
6
I
=
If
×
(1
−
e
−
t
÷
τ
)
1.264
=
2
×
(1
−
e
−0.5 ÷ 0.5
)
U
=
1
2
×
L
×
If
2
6
=
1
2
×
3
×
2
2
Assumptions
•
The switch closes at t = 0 with no current; the battery and wires are ideal, and R is the whole loop’s resistance.
•
Loop rule: ε = IR + L dI/dt, solved by I = If(1 − e−t/τ) with the time constant τ = L ÷ R (H ÷ Ω gives s).
•
The coil’s back emf at first stops the current; it settles at If = ε ÷ R, when the coil drops no voltage.
•
The coil then stores U = ½LIf² in its magnetic field.
Step-by-step
WE KNOW
ε = 12 V, R = 6 Ω, L = 3 H, t = 0.5 s
FIND
time constant (τ), final current ε ÷ R (If), current at t (I), energy stored at the end (U)
Step 1 · Find time constant (τ)
Use τ = L ÷ R
The
time
constant
is
L
over
R
:
henries
over
ohms
gives
seconds.
τ
=
3
6
τ
=
0.5
s
Step 2 · Find final current ε ÷ R (If)
Use If = ε ÷ R
Once
the
current
is
steady,
dI/dt
=
0
and
the
coil
drops
no
voltage,
so
all
of
ε
is
across
R
.
If
=
12
6
If = 2 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
=
2
×
(1
−
e
−0.5 ÷ 0.5
)
I
=
2
×
(1
−
e
-1
)
I
=
2
×
(1
−
0.3679)
I
=
2
×
0.6321
I
=
1.264
A
Step 4 · Find energy stored at the end (U)
Use U = ½ × L × If²
The
work
done
against
the
back
emf,
the
integral
of
LI
dI
from
0
to
If,
is
1
2
LIf²,
stored
in
the
field.
U
=
1
2
×
3
×
2
2
U
=
1
2
×
3
×
4
U
=
1.5
×
4
U
=
6
J
Check
Put the numbers back in. Both sides should match.
0.5
=
3
6
✓
2
=
12
6
✓
1.264
=
2
×
(1
−
e
−0.5 ÷ 0.5
)
✓
6
=
1
2
×
3
×
2
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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