Exercise 31-59: Demonstrating inductance
A demonstration of inductance that is often used in physics courses employs a circuit
such as the one shown below. Switch S is closed, and the light bulb, represented by
resistance R1 , is seen to just barely glow. After a period of time, switch S is opened, and
the bulb is then seen to light up brightly for a short period of time. This effect is easy
to understand if one thinks of an inductor as a device that imparts an ”inertia” to the
current, preventing a discontinuous change in the current through it.
1. Derive, as explicit functions of time, expressions for i1 and i2 , the currents through
the light bulb and inductor, after switch S is closed.
2. After a long period of time, steady-state conditions can be assumed. Obtain expressions for the steady-state currents in the bulb and the inductor.
3. Switch S is now opened. Obtain an expression for the current through the inductor
and light bulb as an explicit function of time.
4. You have been asked to design a demonstration apparatus using the circuit shown
in the figure with a 52.0-H inductor and a 60.0-W light bulb. You are to connect a
resistor in series with the inductor, and R2 represents the sum of that resistance plus
the internal resistance of the inductor. When switch S is opened, a transient current
is to be set up that starts at 0.900 A and is not to fall below 0.300 A until after
0.200 s. For simplicity, assume that the resistance of the light bulb is constant and
equals the resistance the bulb must have to dissipate 60.0 W at 120 V. Determine
R2 and ε for the given design considerations.
5. With the numerical values determined in the previous part, what is the current
through the light bulb just before the switch is opened? Does this result confirm
the qualitative description of what is observed in the demonstration?
1
Derive expressions for i1 and i2
Kirchoff tells us that ”the algebraic sum of the potential differences in any loop must
equal to zero.” Applied on this circuit does it give the following result:
ε − i2 R2 − L
di2
=0
dt
When solving this, we should keep in mind that the intial and final currents are given by
ε
i2 t=0 = 0 and i2 t=∞ =
,
R2
To solve the current i2 , all terms with i are moved to the right and all terms with t
to the left.
di2
ε − i2 R2
i2 R2 − ε
=
=−
dt
L
L
1
di2
= − dt
i2 R2 − ε
L
Z t
Z i2
R2
di2
=
dt
−
i
−
ε/R
L
2
2
0
0
R2
i2 − ε/R2
= − t
ln
−ε/R2
L
R2
R2
−i2
+ 1 = e− L t
ε
R2
ε i2 =
1 − e− L t .
R2
The current through the resistor R1 can be found by applying Ohm’s law:
i1 =
2
ε
R1
Obtain expressions for the steady-state currents
The current through resistor R1 stays the same, since the voltage across the resistor
doesn’t change. When a steady-state current flows through the inductor, no ε is induced,
so the inductor can be seen as a wire now.
i2 =
3
ε
ε
, i1 =
R2
R1
Obtain an expression for the currents when S is
opened
Now the circuit consists only of the two resistors and the inductor in series, so the currents
i1 and i2 are equal. Kirchoff’s law gives us now:
iR + L
di
=0
dt
Now solve it, keeping in mind that the current at t = ∞ is zero.
iR = −L
di
dt
R
di
dt =
L
i
Z t
Z i
R
di
− dt =
L
0
i0 i
Rt
−
= ln i − ln i0
L
−
R
i = i0 e− L t
Here i0 is the current at t = 0 and that’s when S gets closed, so i0 = ε/R2 . The resistance
R of the circuit is now the sum of the two R = R1 + R2 . This results in the following
expression for i:
ε − R1 +R2 t
L
i=
e
R2
4
Determine R2 and ε for the given design considerations
First, compute the resistance R1 of the light bulb. The resistance is equal to the resistance
at 120V, 60W. Under these circumstances, the current through the light bulb is I =
P/V = 60/120 = 0.5A. With Ohm’s law we get R = V /I = 120/0.5 = 240Ω.
The current at t = 0 is equal to 0.9A, so ε/R2 = 0.9A. Using the result of the
previous question, we can derive another relation between ε and R2 . At t = 0.2s the
current i = 0.3A:
240+R2
0.3A = 0.9 e− L 0.2
With L = 0.59H this gives R2 = 45.6Ω and ε = 41.1V.
5
Compute the current through the light bulb before
S is opened
The current through the light bulb i1 = ε/R1 = 41.1/240 = 0.17A. This is much less than
the 0.3A that had to go through the light bulb in the previous question. The light bulb
will probabely just barely glow or not at all. This confirms the qualitative description at
the beginning of the question.
This document was created by Willem van Engen (wvengen@stack.nl)
Homepage: http://willem.n3.net