Transcript Chapter 24-

Chapter 30--Examples
1
Problem
In an L-C circuit, L=85 mH and C=3.2 mF.
During the oscillations the maximum
current in the inductor is 0.85 mA
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 a magnitude
of 0.5 mA?
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i=-wQsin(wt+f)

At maximum, sin()=1, so
 i=-wQ
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Where Q is the maximum value
w=
1
1

 1917 rad 1
LC
(85E  3)(3.2 E  6)

Q=0.85e-3/1917
 Q=0.44 mC
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(½ Li2)+(q2/2C)=(Q2/2C)

We know
L=85 mH
 C=3.2 mF
 Q=4.4e-7 C

We need to solve for q when i=0.5mA
 q=3.58e-7 C

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Problem

1.
2.
In the circuit to the right,
EMF=60 V, R1=40 W, R2=25
W, and L=0.30 H
Switch S is closed. At some
time t afterward, the current
in the inductor is increasing
at a rate of di/dt=50 A/s. At
this instant what is the
current thru R1 and R2 ?
After the switch has been
closed a long time, it is
opened again. Just after it
is opened, what is the
current thru R1?
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The inductor acts to oppose the
source creating the change

So we can replace the inductor with an
EMF device, wired to oppose the 60 V
device with an EMF= L*di/dt =0.3*50=15
V
60 V
40 W
25 W
15 V
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Using the loop rules
 I1=60V/40 W  1.5 A
 I2=(60-15)/25 W 1.8

A
Part B)
After a long time, the current becomes
steady, and di/dt=0 so I2=60/25=2.4 A
 At the instant the switch is opened, the
inductor maintains this current so I2=I1=2.4
A

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Problem

In the circuit, switch
S is closed at t=0.
a) Find the reading of
each meter just
after S is closed.
b) What is the reading
of each meter along
time after S is
closed?
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Part B) First!
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After a long time, the inductors act like regular
wire.
So we add the 5, 10, 15 W resistors in parallel
to get 2.72 W
itotal =25 V/ (40+2.72 W )=0.585 A so A1=0.585
A
The voltage across the parallel circuit is
0.585*2.72=1.57 V
A2=1.57/5 = .314 A
A3=1.57/10=.157 A
A4=1.57/15=.104 A
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Part A– The conceptual thing

At the very instant that S is closed,
inductors are maximally resisting. So
they essentially PREVENT current flow
through the 5 W and 10 W resistor!
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A2=A3=0
Thus A1 must equal A4 which is 25 V/
(40+15 W) =0.455 A
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Problem
In the circuit, the switch S has
been open for a long time
and is suddenly closed.
Neither the battery nor the
inductors have any
appreciable resistance.
a) What do the ammeter and
voltmeter read just after S is
closed?
b) What do the ammeter and
the voltmeter after S has
been closed a very long
time?
c) What do the ammeter and
the voltmeter read 0.115 ms
after closing S?
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Part A)
At the very instant the switch is closed,
the inductors act to stop current flow so
A=0
 However, the potential difference is the
full 20 V.

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Part B)
Again, after a long time, the inductors
just act like wires so
 i=20 V/(50+25 W) =0.267 A
 V=0 since no potential difference.

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Part C)

To add the inductors, note that the 12 mH and
15 mH inductors are in series so L1215=27 mH
 L1215 is in parallel with the 18 mH so
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1/LTot= (1/18)+(1/27)
LTot=10.8 mH
RTot=75 W
i=(EMF/RTot)*(1-e-(R/L)*t)
i=(20/75)*(1-e-(75/10.8e-3)*0.115e-3)=0.147 A
VR =iR=0.147*75=11 V
So potential difference across the inductor
network must be 20-11= 9V
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