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Charging of a capacitor
• One of the functions of a capacitor is its ability to store
up charge when a potential difference is applied across
the positive and negative plates.
• Energy is stored in the electric field. When a voltage is
applied across a capacitor, current rushes into the plates
of the capacitor, developing a potential difference across
the capacitor, therefore the potential difference between
the battery and the capacitor become smaller and the
flow rate of electrons become smaller.
• The charging process continues until the capacitor
becomes fully charged. The charging current follows an
exponential curve.
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Charging of a capacitor…/2
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Charging of a capacitor…/3
q
q
E  VR  VC ,
C
,
VC 
VC
C
q
dq
E  iR  , EC  RC
q
C
dt
dq
EC  q  RC
dt
dt
dq
 RC   EC  q ,
dt
d EC  q 
 RC   EC  q
t
C1 
  ln EC  q ,
RC
EC  q  e
C1  Integration Cons tan t
C1 t / RC
EC  q  C2 e t / RC ..................equation
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Charging of a capacitor…/4
Insert initial (boundary) conditions, if the capacitor is
initially uncharged, then at t = 0, q = 0
C2 = EC, substitute into equation 1
EC  q  ECet / RC

q  EC 1  e
 t / RC

 t / RC
q
t / RC


V

E
1

e

VC   E 1  e
 C
C
dq
d
i
 EC 1  e t / RC
dt
dt
1 
E t / RC

t / RC
i  EC  0  e


i

e

RC 
R


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Charging Curve

q
VC   E 1  e t / RC
C
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Charging Curve

q
VC   E 1  e t / RC
C

• From the charging equation, we noticed that the
rate of charging is determined by the exponential
curve that is in term determined by the RC constant.
• The term RC is termed the time constant since it
affects the rate of charge. Mathematically, this is
the time taken for the capacitor to reach 0.632 of
the fully charged value.
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Duration of Transient
• Theoretically, capacitors will never be fully charged
according to the charging equation.
• Thus, for all practical purposes, transients can be
considered to last for only five times of the time
constant. I.e. the capacitor is said fully charged after
5× RC.
• After 5 time constant, q, Vc and I will be over 99% to
their final values.
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Charging of Capacitor-Example one
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Charging of Capacitor-Example one
• An uncharged capacitor of 2000 micro-Farad is
connected to a 100 volt D.C. supply in series
with a current limiting resistor of 5000 Ohms,
calculate
• i) The voltage of the capacitor at the end of 8
seconds charging
• ii) The charging current at the end of 8 seconds
• iii) The time taken for the capacitor to be
charged to 80 volts.
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Charging of Capacitor-Solution to Example one
1) Using the charge formula, VC  q  E 1  et / RC 

VC  100 1  e
C
8 / 5000 200010 6

VC  55.07 Volts
2)
E  t / RC
i e
R
100 8 / 5000200010
i
e
5000
6
i  8.9866 milli  Ampere
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Charging of Capacitor-Solution to Example
one…/2
3)
For the capacitor to be charged to 80 Volts, using
the same formula, VC  q  E 1  et / RC
C

80  100 1  e


t / 5000200010 6

0.8  1  et / 10
et / 10  0.2
 t / 10  ln 0.2  1.6094
t  16.094 Seconds
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Charging of a initially charged capacitor
• If at the start of charging, the capacitor is charged to a
voltage of E1 Volts,
• Then at t = 0, q = E1, substituting this initial condition
 t / RC
into equation 1, we have EC  CE1  C2 e
C2  CE  E1 
EC  q  C E  E1  e t / RC
q  EC(1  e
 t / RC
 t / RC
)  E1Ce
q
VC   E  Ee t / RC  E1e t / RC
C
VC  E (1  e  t / RC )  E1e  t / RC
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Charging of a initially charged capacitor
dq
d
d t / RC
t / RC


i
 EC 1  e
 E1C e
dt
dt
dt
E  E1  e t / RC
E
E
i  e t / RC  1 e t / RC  i 
R
R
R
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Example to charging of a capacitor with
residual charge and initial voltage
A 3,000μF capacitor has an initial voltage of 50 Volts is
further charged by a 200 volts D.C. supply in series with
a 2 k-Ohm resistor. Calculate the voltage across the
terminals of the capacitor after 10 seconds.
Solution: Using the formula for capacitor with an initial
t / RC


V

E

E

E
e
voltage, C
1
VC  200  50  200e
10 / 2000300010 6
VC  200  150  0.1889
VC  171.67 Volts
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Alternative solution to previous problem
• Alternatively, if you do not wish to memorize the
formula for charging a capacitor with an initial charge,
you can first find the time required to charge an
uncharged capacitor to the initial voltage, then add the
extra time for charging to find the final voltage.
Solution:
Time required to charge an uncharged 3,000μF
capacitor to 50 volts can be found by applying the
formula
 t / RC

VC  E 1  e
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Alternative solution to previous problem…/2

t / 2000300010 6
50  200 1  e

50
 1  et / 6
200
0.75  et / 6
 0.2877  t / 6
t  1.7261 Seconds
Total time equivalent the capacitor is to be charged from zero
volt = 1.7261+ 10 = 11.7261 Seconds
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Alternative solution to previous problem…/3
Final voltage of the capacitor =

11.7261 / 2000300010 6
VC  200 1  e

1.9544
VC  200 1  e


VC  200 0.8583
VC  171.67 Volts
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Discharge of a capacitor
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Discharge of a capacitor…/2
0 = VR +VC
0  iR 
q
C
q
 iR
C
dt
dq


RC
q
t

 C1  ln q
RC
C1  Cons tan t
t / RC  C1
qe
 t / RC
q  C2e
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C2  Cons tan t
Chapter 2.3 Capacitor Charging & Discharging Page 19 of 23
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Discharge of a capacitor…/3
Substitute boundary condition, at t = 0,
Voltage across C = E, q = EC
C2 = EC
q  ECet / RC
q
VC   Eet / RC
C
dq
1 t / RC
i
 EC  
e
dt
RC
E  t / RC
i e
R
(Note that the negative sign indicate that the current is
opposite to the charging current’s direction)
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Discharge curve
E t / RC
i e
R
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Discharge of capacitor example
A 1000μF capacitor previously charged to 80 Volts is to
be discharged through a resistance of 20 k-Ohms. Find
the voltage across the terminals of the capacitor at the
end of 15 seconds.
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Discharge of capacitor-solution to example
Using the discharge formula,
q
 t / RC
VC   Ee
C
VC  80e
15 / 20000100010 6
VC  37.79 Volts
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