Energy stored in capacitor
V = VB
V = VB
Charging capacitor
Stored charge
+Q
+Q
V = VB
V = VB
-Q
-Q
V = VB
Stored charge produces the current
+Q
-Q
10mA
V = VB
R
Energy stored in capacitor
Electric Energy = Charge x Voltage: W = Q × V
This formula would be true for capacitors, had the
charge between plates been transferred at a constant voltage.
For capacitors, Q = C × V
As the charge Q on the plate increases,
the voltage V increases too (and vice versa).
Suppose the capacitor was charged to the voltage V0:
Q
The stored energy is the area under
the Q – V line.
Q0
Note that, Q0 = C V0:
WC
V0
V
V0 Q0
WC =
2
CV02
WC =
2
V0 is the voltage on the charged capacitor
Example problem 1
The capacitor of 1 mF has been charged to 100 V.
What energy is stored in the capacitor in Joules?
WC
2
CV0
=
2
0
of
5
120
Timed response
Commercial capacitors
C=
20 mF capacitor
ε d ⋅ ε0 A
d
Capacitor bank (series-parallel)
Parallel connection of capacitors
V = VB
C1
C2
Q1 = C1 V1
Q2 = C2 V2
Capacitors C1 and C2 are connected in parallel: both terminal of each
capacitor are connected to the same wires.
The voltage on each of the capacitors is the same, VB
The charge on the capacitor C1, Q1 = C1×VB
The charge on the capacitor C2, Q2 = C2×VB
Total charge stored in both capacitors:
QTot = Q1 + Q2 = (C1 + C2) × VB
The equivalent capacitance, Ceq = QTot/ VB = C1 + C2
Cpar= C1 + C2
Parallel connection of capacitors
C1
V = VB
IT
C2
I1
I2
We can also find the equivalent capacitance from the KCL.
For the 1st and 2nd capacitors,
∂V
∂V
I1 = C1
∂t
; I 2 = C2
∂t
According to the KCL, the total current IT = I1 + I2.
Substituting the values for I1 and I2:
 ∂V
∂V 
∂V
∂V
 = (C1 + C2 )
+ C2
= C par
IT =  C1
;
∂t
∂t 
∂t
∂t

Cpar= C1 + C2
;
Series connection of capacitors
V = VB
C1
C2
Q = C1 V1 ; Q = C2 V2 ;
V1
V2
C1
V2 = V1
C2
Capacitors C1 and C2 are connected in series:
the charge Q on C1 and C2 is the same; For each of the capacitors,
From the KVL, VTot= V1 + V2.
The total charge, on the C1 – C2 combination is still Q.
The equivalent capacitance is defined as: Q = CEqS VTot

C1 
Q = CEqS (V1 + V2 ) = CEqS  V1 + V1  ; Q = C1 V1 ;
C2 

1/CSer= 1/C1 + 1/C2

CC 
 CS = 1 2 
C1 + C2 

 C1 
CEqS 1 +  = C1 ;
 C2 
Example problem 2
Three capacitors 2 mF each are connected in parallel.
What is the total capacitance in mF?
0
of
5
120
Timed response
Example problem 3
Three capacitors 2 mF each are connected in series.
What is the total capacitance in mF?
0
of
5
120
Timed response
Example problem 4
Parallel-plate capacitor has a capacitance of 9 nF.
A thin metal plate has been inserted in the middle
between the top and bottom plates.
What is the capacitance (in nF) of the capacitor now?
C=
ε d ε0 A
d
0
of
5
120
Timed response
Transients in R-C circuit
C
R
VB
d VC
IC = C ×
dt
VR
IR =
R
Series R-C circuit
The first moment after closing the switch, the voltage across the capacitor = 0;
The capacitor behaves as a short-circuit;
The current at t=0, I0 = VB/R;
After all the transients are over (t Yh) , I = 0
Commutation rule for capacitors
C
VB
R
VC
V_
Consider a capacitor right before and right
after commutation in an arbitrary circuit.
The capacitor voltage (charge) does not
have to be zero before the commutation.
V+
Commutation event
time
If VC changes instantaneously after the commutation, the current in the
connected circuit would be infinitely high:
d VC
V+ − V−
IC = C ×
=C×
dt
dt |→0
If V+ is different from V- when dt 0,
then IC ∞
The capacitor voltage does not change after commutation: VC- = VC+
Graphs showing the current and
voltage for a capacitor charging
VB
I (t ) = e
R
−
t
RC
Capacitor voltage
t
−


RC
VC (t ) = VB × 1 − e



τRC = R×C
When t = 3 ×τRC,
VC = 0.95VB ;
Graphs showing the current and
voltage for a capacitor discharging
Capacitor starting voltage is VB
VB
I (t ) = e
R
−
t
RC
VC (t ) = VB e
−
t
RC
τRC = R×C
When t = 3 τRC,
VC = 0.05VB=5% VB
General formula for step response of an arbitrary R-C circuit
C
R
VS
R-C circuit
vC = vCF + ( vC 0 − vCF ) e
−t /τ
τ = RC
VC0 is the capacitor voltage right after (or right before) the commutation;
VCF is the capacitor voltage long time after all the transient processes are over.
R is the total resistance connected to the capacitor after commutation
(al the sources are zeroed to find the equivalent total resistance)
Example 1
Delayed alarm circuit
S1
R=10 k
VB=4.5 V
S2
C=0.5 mF
Sensor switch: S1.
Electronic switch S2 triggers the alarm system when the voltage
across it exceeds the preset threshold value VT.
R = 10k; C = 0.5 mF; VB = 4.5 V.
Assume the S2 resistance infinitely high.
The required time delay between the switch S1 turn-on and triggering
switch S2 must be tt = 3s.
What threshold voltage VT must the switch S2 be tuned to?
Example 1
Delayed alarm circuit
S1
R=10 k
VB=4.5 V
S2
C=0.5 mF
vC = vCF + ( vC 0 − vCF ) e − t / τ
VC0 = 0; VCF = VB; τ = R×C;
vC = vB − vB e − t / τ
The required alarm triggering time: tt=3s
The required switch threshold voltage vSW = vc(tr):
vsw = 4.5 − 4.5 e −3 /(10 e 3×0.5e−3) = 4.5 − 4.5 × e −0.6 = 2V
Example 2
The switch in the circuit shown in Fig. 7.25 has been in position
a for a long time. At t = 0 the switch is moved to position b.
What is the vC time dependence at t>0?
vC = vCF + ( vC 0 − vCF ) e − t / τ
vC0 = V60Ω = −40V *60/(60+20) = -30 V
vF = 90 V;
τ = RC = 400 kΩ ∗ 0.5µF = 0.2 s.
vC = 90 −120 e − t / 0.2 V