# Physics 6B Inductors and AC circuits Prepared by Vince Zaccone For Campus Learning

```Physics 6B
Inductors and AC circuits
Prepared by Vince Zaccone
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Inductance
Mutual Inductance of two coils:
Some of the magnetic flux through one
coil also passes through the other coil,
inducing a voltage.
Inductance is magnetic flux/current.
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Self-Inductance
Changing current through the wires in a
coil will induce a voltage that opposes
the CHANGE in the current.
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R-L Circuit
When connected in a circuit with a
resistor, an inductor will have the
effect of slowing down changes in the
current through the resistor.
When the current is steady (the switch
has been closed for a long time), the
inductor has no effect, but there is
potential energy stored in the inductor.
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R-L Circuit
If switch S1 is closed in the circuit,
current will begin to flow through the
resistor and inductor as shown. This
increasing current will induce current
to flow the opposite direction, slowing
the growth of the current.
We can write down a formula for the
current as a function of time:
π π‘ =
π
1 − π−
π
π
πΏ π‘
πΏ
The quantity π = π is called the “time
constant” for this exponential decay.
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R-L Circuit
Once the current reaches a steady
value we can flip the switches,
opening S1 and closing S2. Then
current will keep flowing for while as
the inductor opposes this decreasing
current.
A similar formula describes this
decaying current as a function of time:
π π‘ = πΌ0 π −
π
πΏ π‘
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Example: A 35.0V battery with negligible internal resistance, a 50.0Ω resistor and a 1.25mH
inductor are connected in series with an open switch. The switch is suddenly closed.
(a) How long after closing the switch will the current through the inductor reach half of its
maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach half its
maximum value?
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Example: A 35.0V battery with negligible internal resistance, a 50.0Ω resistor and a 1.25mH
inductor are connected in series with an open switch. The switch is suddenly closed.
(a) How long after closing the switch will the current through the inductor reach half of its
maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach half its
maximum value?
As soon as the switch is closed, current begins to flow around the circuit, increasing toward
a maximum value given by Ohm’s Law. Here is the formula:
π π‘ =
π
1 − π−
π
π
πΏ π‘
We want to find the time when the current is half of the maximum.
π π‘ =
1
2
π
π
=
1 − π−
π
π
− π πΏ π‘ = ln
1
2
π
πΏ
→π‘=−
β ln
π
πΏ π‘
1
2
1
2
→ = 1−
π
π− πΏ π‘
1.25 β 10−3 π»
=−
β ln
50Ω
1
2
→
π
π− πΏ π‘
1
=
2
= 1.73 β 10−5 π  = 17.3ππ
For part b) we want the energy to be half of its maximum, so use the energy formula:
π = 12πΏπ 2 = 12 12πΏπΌ 2 → π(π‘) =
πΌ
2
Using the formula for current again:
1
2
= 1 − π−
π
πΏ π‘
→ π‘ = 30.7ππ
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L-C Circuit
A circuit containing a capacitor and an inductor will exhibit an oscillating
current, with potential energy transferring back and forth.
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L-C Circuit
The oscillations in an L-C circuit should
look familiar. This situation is directly
analogous to an undamped mass-spring
system that we saw previously.
All of the formulas we developed for that
case are repeated here, with charge, q,
taking the place of displacement, x. The
capacitor is related to the spring constant,
and the inductance is like mass.
To add in the damping, we just include a
resistor in the circuit…
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
(a) What is the maximum charge on the capacitor?
We can use energy for this if we want to. When all the energy is in the inductor it will have
maximum current. When all the energy is in the capacitor it will have maximum charge.
ππππ = 12πΏπΌ 2 = 12(85 β 10−3 π»)(0.85 β 10−3 )2 = 0.307ππ½
πππππ =
2
1π
2 πΆ
= 0.307ππ½ → π = 44.3ππΆ
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
b) We can use energy again, or we can use the formula for the charge as a function of time.
πΈπ‘ππ‘ππ =
1 2
πΏπ
2
+
2
1π
2 πΆ
Total energy can be found from max current or max charge. Should be the same either way.
We can solve for the charge when the current is as given:
π2
1
1
−8
−3
−3
2
πΈπ‘ππ‘ππ = 3.07 β 10 π½ = 2 85 β 10 π» 0.5 β 10 π΄ + 2
→ π = 3.58 β 10−7 πΆ = 358ππΆ
3.2 β 10−6 πΆ
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
To go from no charge to fully charged is a quarter of a cycle, so we need to find the period of
the oscillation. We have a formula for angular frequency:
π=
1
= 1917πππ
π
πΏπΆ
Rearrange this to get the period, then divide by 4:
π=
2π
= 0.00328π  → 14π = 0.82ππ
π
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L-R-C Series Circuit
The AC voltage source drives the oscillating
current in the circuit. Frequency of the source will
affect the overall impedance (Z) of the circuit.
π=
1
− ππΏ
ππΆ
2
+ π2
At resonance the impedance will be smallest, and the current will be
largest. This happens when the capacitor and inductor terms balance out.
We can solve for the resonant frequency:
ππΏ =
1
→ ππππ  =
ππΆ
1
πΏπΆ
Notice that at this frequency the impedance just equals the resistance R.
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L-R-C Series Circuit
If there is a driving voltage V(t) our equation for the circuit looks like this:
βπ(π‘)
π π‘
πΏ
+ π(π‘)π +
= π(π‘)
βπ‘
πΆ
Solving this equation gives a general solution for i(t):
π(π‘) =
π=
π0
π ππ ππ‘ + π
π
1
− ππΏ
ππΆ
π = arctan
2
+
π2
1 1
− ππΏ
π ππΆ
Z is called IMPEDANCE. It can be used
in place of resistance for an AC circuit.
This is the phase angle. You can use it
to determine whether the circuit
voltage leads or lags the current.
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