Transcript Chapter 24--Examples 1

Chapter 24--Examples
1
Problem
In the figure to the left, a
potential difference of
20 V is applied across
points a and b.
a) What is charge on
each capacitor if C1=
10 mF, C2=20 mF, and
C3=30 mF.
b) What is potential
difference across
points a and d?
c) D and b?
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Step 1: Find Equivalent Resistance
C1&C2: 10+20=30
 C1&C2+C3: (1/30)+(1/30)= 2/30 i.e.
30/2=15 mF

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Finding Potentials
Q=CV=15*20=300 mC
 So charge on C3 is 300 mC
 The voltage across C3 is 300 mC/30 mF=
10 V
 So 10 V across points b & d and
therefore, 20-10=10 V across a & d

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Parallel Network

If there is 10 V across this network then
Q1=10 mF*10V= 100 mC
 If there are only 300 mC total and 100 mC is
on this capacitor, then 200 mC must be
charge of C2
 Check: Q2=20 mF*10V=200 mC

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Problem
In the figure, each
capacitor C1=6.9 mF and
C2=4.6 mF.
a)Compute equivalent
capacitance between a
and b
b)Compute the charge on
each capacitor if Vab=420 V
c)Compute Vcd when
Vab=420 V
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Analysis
Look at the right most connection: it is
parallel connection between the C2
capacitor and the 3 C1 capacitors
 3 C1’s: 6.9/3=2.3 mF
 C2+3C1’s=2.3+4.6=6.9

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Repeat Again
Now the 6.9 mF capacitor is in series
with the other C1 capacitors
 So it is the same circuit as again, so the
equivalent is 6.9 mF
 Finally, the total equivalent is 2.3 mF

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So 420 V *2.3 mF= 966 mC
For each C1 capacitor on the leftmost
network, the voltage across each is 140
V (966/6.9 or 420/3)
 If there is 140 V across the C2, then
140*4.6=644 mC
 There must be 966-644 = 322 mC in the
other branch.

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In the middle network,
Each capacitor has 46.67 V (140/3 or
322 mC/6.9)
 So voltage across c & d is 46.67 V
 Then C2 capacitor has 214mC and the
other C1 capacitors have 322-214 =107
mC

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Problem
Two parallel plates have equal and
opposite charges. When the space
between them is evacuated, the electric
field is 3.2 x 105 V/m. When the space
is filled with a dielectric, the electric
field is 2.5 x 105 V/m.
a) What is the charge density on each
surface of the dielectric?
b) What is the dielectric constant?
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ETotal=E0-Ei
++++++++++++++++++++++
-- - - - - - - - - - - - - - - - - Ei
+++++++++++++++++++
E0
E  ETotal
i
 Ei 
0
 i   0 * ( E  ETotal )
----------------------i=8.85e-12*(3.2-2.5)*105=4.32x106 C/m2
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Dielectric Constant

K=E0/E=3.2e5/2.5e5=1.28
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Problem
A 3.4 mF capacitor is initially uncharged and then
connected in series with a 7.25 kW resistor and
an emf source of 180 V which has negligible
resistance.
a) What is the RC time constant?
b) How much time does it take (after connection)
for the capacitor to reach 50% of its maximum
charge?
c) After a long time the EMF source is
disconnected from the circuit, how long does it
take the current to reach 1% of its maximum
value?
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RC Value
RC=3mF*7.25kW
 RC=3e-6*7.25e3
 RC=21.75 ms

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Time to 50% of max charge
Q(t)=C*V*(1-e-t/RC)
 Q(t)/CV is the fraction of the maximum
charge so let Q(t)/CV =50%
 .5=1-e-t/RC
 e-t/RC=.5=1/2 or 2-1
 -t/RC=-ln(2)
 t=RC*ln(2)=21.75*.693
 t=15.07 ms

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Since R & C have not changed,
RC=21.75 ms
I(t)=(V/R)*e-t/RC
 I(t)/(V/R) is the fraction of maximum
current
 Let I(t)/(V/R) = 1% or 0.01
 0.01=e-t/RC
 ln(0.01)=-4.605=-t/RC
 t=21.75*4.605=100 ms

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