Transcript Energy stored in capacitors File

Energy in a Capacitor
Energy in a Capacitor
The energy stored in a
capacitor can be used to
temporarily drive a circuit.
This is in full analogy with
the potential energy of water
stored in an upper reservoir.
This water energy can be
used to drive a water mill.
Energy in a Capacitor
As a capacitor is charged using
constant current, I =Q/t for time
t, the voltage increases from 0 to
V. So the average voltage is ½V.
Power is P=IV, energy is
E=Pt=IV x t, so the total energy
delivered to the capacitor is:
E  I  12 V  t
 (Q / t )  1 2 V  t 
1
2
QV
Energy in a Capacitor –
different equations for energy
Can you solve
Q
1
E in terms of
E  2 QV AND C  for
V & C, and then
V Q & C?
But Q=CV, thus, substituting to the equation
above we derive
2
E= ½QV = ½CV
On the other hand, V=Q/C, thus, substituting
to the equation E= ½QV we derive
E= ½QV = ½Q2/C
Energy in a Capacitor – Which
equation to use?
The energy stored
in a capacitor is
E = ½QV
= ½CV2
= ½Q2/C
E = ½ CV2= ½ 10 [mF] x
202 [V2] = 2000 mJ
Note that if we halve the
voltage, the energy
decreases four times:
E = ½ CV2= ½ 10 [mF] x
102 [V2] = 500 [mFxVxV]=
500 [mCxV] =500 mJ
Problem:
A 10 mF capacitor is
charged to 20 V. How
much energy is stored?
Think what you know
and what you do not
know.
We know the
capacitance and the
voltage, but we do not
know the charge, so we
should use:
E= ½CV2
Energy in a Capacitor
The energy stored in a
capacitor can be used to
temporarily drive a circuit.
As a capacitor is charged
using constant current, I
=Q/t for time t, the
voltage increases from 0 to
V. So the average voltage
is ½V.
Power is IV, energy is
IV x t, so the total energy
delivered to the capacitor
is: E  I  1 V  t
2

1
2 QV
The energy stored
in a capacitor is
E = ½QV
= ½CV2
= ½Q2/C
Capacitors in Series -Q
If two capacitors C1 and C2
are connected in series, each +Q
capacitor must store the same -Q
charge, so voltages on the
+Q
capacitors are:
V=V1+ V2
V1=Q/C1
V2=Q/C2
Q
Q But, C=Q/V,
V1  ; V2 
C1
C 2 thus, V=Q/C
Also (the second Kirchoff’s But, C=Q/V,
Q Q thus,
law):
V  V1  V2  
C1 C 2 1/C=V/Q
1 V Q Q 1
1
1
       
So
C Q  C1 C2  Q C1 C2
Capacitors in
series add in
reciprocal.
1/C=1/C1+1/C2
Capacitors in Series
If two capacitors C1 and
C2 are connected in
series, each capacitor
must store the same
charge, so:
Q
V1 
C1
Q
V2 
C2
Also V  V1  V2  Q  Q
C1 C 2
So
1 V Q Q 1
  

C Q  C1 C 2  Q
1
1


C1 C 2
Capacitors in
series add in
reciprocal.
1/C=1/C1+1/C2
Capacitors in Parallel
Capacitors in parallel must have
the same potential difference
across them: V1=V2=V (the
A
-Q1
V1= V -Q2 V2= V
second Kirchoff law for nodes A and B).
Charge stored on the 1st and 2nd +Q1
capacitors are Q1=C1V1=C1V But, C=Q/V,
and Q2=C2V2=C2V.
thus, Q=CV
Total Charge Q=Q1+Q2 stored,
Q  Q1  Q2
 C1V  C2V
Combined capacitance,
Q C1V  C 2V
C 
 C1  C 2
V
V
Capacitors in parallel add
B
+Q2
Capacitors in
parallel add.
C=C1+C2
Capacitors in Parallel
Capacitors in parallel must
have the same potential
difference across them.
Total Charge stored,
Q  Q1  Q2
 C1 V  C 2 V
Combined capacitance,
Q C1 V  C 2 V
C 
V
V
 C1  C 2
Capacitors in parallel add
Capacitors in
parallel add.
C=C1+C2
Capacitors in more complex parallel-series circuits
•Capacitors with
capacitances C1=24 mF,
C2=4 mF, and C3=8 mF are
connected in series.
•These three capacitors
can be replaced by one
effective capacitor with
Ceff-1
• This effective capacitor
(Ceff-1) is in parallel with
C4= 4 mF.
•These two capacitances
(Ceff-1 and C4) can be
replaced by one effective
capacitance Ceff-2
• The blue capacitor block
with Ceff-2 is in series with
C5=5 mF and C6= 8 mF
C5
C4
C6
C1
C3C CCeff-1
eff-2 2
Capacitors in more complex parallel-series circuits
•Consider three Capacitors
with capacitances C1=24
mF, C2=4 mF, and C3=8 mF
which are in series.
Capacitors in series
add in reciprocal.
1/C=1/C1+1/C2
• 1/Ceff-1=1/C1+1/C2+1/C3
= 1/24+1/4+1/8
=10/24=5/12 [1/mF]
• Therefore, the effective
capacitance of the red
box is Ceff-1=12/5 = 2.4
mF
C5
C4
C6
C1
C3
Ceff-1
C2
Capacitors in more complex parallel-series circuits
•Consider two Capacitors
with capacitances Ceff-1
=2.4 mF, C4=4 mF which
are in parallel.
Capacitors in parallel
add.
C=C1+C2
• Ceff-2=Ceff-1+C4 = 2.4+4.0
=6.4 mF
Ceff-2
C5
C6
C4
Ceff-1
2.4 mF
Capacitors in more complex parallel-series circuits
•Consider three Capacitors
with capacitances C5=5
mF, Ceff-2=6.4 mF, and
C6=8 mF which are in
series.
• 1/Ceff-total =1/C5 +1/C6
+1/Ceff-2 = 1/5+1/6.4+1/8
= 0.48 [1/mF]
• Therefore, the effective
capacitance of the red
box is Ceff-total= 2.1 mF
Capacitors in series add
in reciprocal.
1/C=1/C1+1/C2
C5
Ceff-2=2.1 mF
C6
6.4 mF
Ceff-2
Capacitors in more complex parallel-series circuirs
• Analyze the circuit and
determine blocks with serial
or parallel connections.
• Using equations for either
parallel or serial connections
calculate effective capacitance
of each block, e.g.,
C5
C6
C4
C1
C3
C2
1/Ceff-1=1/C1+1/C2+1/C3
•Redraw the new circuit
where the blocks are replaced
with the effective capacitance.
•Repeat the same analysis.
C5
C6
C4
1/Ceff-1=1/C1
+1/C2+1/C3
Ceff-1
2.4 mF
Ceff-2=Ceff-1+C4
C5
C6
6.4 mF
Ceff-2
Sharing Charge
A capacitor C1 is charged to Vi
and then disconnected from
the battery.
Then it is connected to
capacitor C2. The voltage on
C1 falls and that on C2 rises as
charge flows from one to the
other. Charge stops flowing
when the voltage on each
equalises at Vf.
I(t)
= Vi
VV11(t)
+Qm+Q(t)
=C1Vi -Q(t)
-Qm
V1(t)
+Q(t) -Q(t)
I(t)
V2(t)
+q(t) -q(t)
Sharing Charge
The voltage V1 on C1 falls and
that (V2) on C2 rises as charge
flows from one to the other (Q
decreases and q increases).
Charge stops flowing when the
voltage on each equalises at Vf.
Two communicating vessels are
the analogy of two sharingcharge capacitors: When one of
the two vessels, which is full of
water, is connected to the other
one a water flux occurs until
water levels in both vessels are
the same.
V1(t)
+Q(t) -Q(t)
I(t)
V2(t)
+q(t) -q(t)
Sharing Charge
When the capacitor C1 is charged (the
voltage V on the capacitor is equal to the
battery emf Vi), it is disconnected from the
V1 = Vi
battery and is connected to capacitor C2.
The voltage on C1 falls from Vi and that on +Q =C V -Q
m
1 i
m
C2 rises as charge flows from one to the
other. Charge stops flowing when the
voltage on each equalises at V1=V2=Vf.
Initial charge (just before disconnection
from the battery which generates voltage
Vi ) equals:
Q  C1Vi
But, C=Q/V,
thus, Q=CV
When two capacitors are connected, the
current between them stops when the
voltage Vf on both capacitors is the same.
Thus, the final charges on capacitors are
Q1  C1V f ; Q1  C 2V
f
Vf
+Q1 -Q1
+Q2
-Q2
Vf
Sharing Charge
Initial charge equals final charge:
Q  C1 Vi  C1 Vf  C 2 V f
So we have the following equation
C1Vi  C1V f  C 2V f
So we can find the final voltage
C1Vi
Vf 
C1  C2 
And Final charges stored on each
capacitor:
When two capacitors
share the charge that
was initially on one, The
resultant voltage is
Vf=C1Vi/(C1+C2)
C1C1Vi
C 2C1Vi
Q1 
; Q2 
C1  C2 
C1  C2 
Sharing Charge
A capacitor C1 is charged to Vi and
then disconnected from the battery.
Then it is connected to capacitor C2.
The voltage on C1 falls and that on
C2 rises as charge flows from one to
the other. Charge stops flowing when
the voltage on each equalises at Vf.
Initial charge equals final charge:
Q  C1 Vi  C1 Vf  C 2 V f
So
C1 Vi  C1 Vf  C 2 Vf
or
C1 Vi
Vf 
C1  C 2 
When two capacitors
share the charge that
was initially on one, The
resultant voltage is
Vf=C1Vi/(C1+C2)