Transcript Lecture 7 - Capacitance

Capacitors in Circuits
Capacitance
Two parallel plates charged
Q and –Q respectively
constitute a capacitor
C=Q/V
-Q
A
E
V
d
+Q
The relationship C = Q / V is valid for any charge configuration
(Indeed this is the definition of capacitance or electric capacity)
In the particular case of a parallel plate capacitor
C = 0 A / d [vacuum] or C =  0 A / d [dielectric]
The capacitance is directly proportional to the area of the plates
and inversely proportional to the separation between the plates
Capacitors in Circuits
+Q
-Q
(Symbol for
a capacitor)
C
V
A piece of metal in equilibrium has a constant value of potential.
Thus, the potential of a plate and attached wire is the same.
The potential difference between the ends of the wires is V,
the same as the potential difference between the plates.
Parallel and Series
Parallel
Series
Capacitors in Parallel
• Suppose there is a potential
difference V between a and b.
• Then q1 V = C1 & q2 V = C2
• We want to replace C1 and C2 with an
equivalent capacitance C = q V
• The charge on C is q = q1 + q2
C 1 - q1
a
b
C 2 - q2
V
a
C-q
• Then C = q V = (q1 + q2 ) V = q1 V + q2 V = C1 + C2
C = C1 + C2
• This is the equation for capacitors in parallel.
• Increasing the number of capacitors increases the capacitance.
b
Capacitors in Series
a
C1
C2
-q +q -q +q
V1
C
b
V2
a -q
+q
b
V
• Here the total potential difference between a and b is V = V1 + V2
• Also V1 = (1/C1) q and V2 = (1/C2) q
• The charge on every plate (C1 and C2) must be the same (in
magnitude)
• Then: V = V1 + V2 = q / C1 +q / C2 = [(1/C1) + (1/C2)] q
• or, V = (1/C) q

1 / C = 1 / C1
+ 1 / C2
• This is the equation for capacitors in series.
• Increasing the number of capacitors decreases the capacitance.
Parallel and Series
Capacitors
Parallel
C = C1 + C2
Series
1/C = 1/C1+1/C2
Parallel
Ceq = C1 + C2 + C3
Series
1/Ceq = 1/C1 + 1/C2 + 1/C3
Parallel and Series
Resistors
Capacitors
1/R=1/R1+1/R2
C=C1+C2
R=R1+R2
1/C=1/C1+1/C2
Parallel
Series
Ideal circuit
Real circuit
How does the capacitor acquire the charge ?
What happens when
the switch is closed ?
RC Circuits: Charging
open
closed
I
R
+
+
V
-
C
V
-
R
VR=IR
+++
- - - VC=q/C
C
V = I(t)R + q(t)/C
When the switch closes, at first a high current flows:
VR is big and VC is small. As q is stored in C, VC increases.
This fights against the battery, so I gradually decreases.
Finally, I stops (I = 0), C is fully charged (VC = Q/C = V),
and Q = C V
RC Circuits: Charging

q  t   C 1  e
 t


q t 
 t
VC  t  
  1 e
C
    t
I t     e
R

  RC  time constant
RC Circuits: Charging

q  t   C 1  e
 t


q t 
 t
VC  t  
  1 e
C
    t
I t     e
R

  RC  time constant
Discharging an RC Circuit
R
C
Open circuit
R
+q
-q
VC=V0
I
C
+q
-q VC=q/C
After closing switch
Current will flow through the resistor for a while.
Eventually, the capacitor will lose all its charge,
and the current will go to zero.
During the transient: q(t) / C – I(t) R = 0
VR=IR
Discharging an RC Circuit
q  t   Qe
 t
q  t  Q  t
VC  t  
 e
C
C
Q
 t
C
I t  
e
R
Charging and Discharging a Capacitor
Charging and discharging of a capacitor occurs gradually
with a characteristic time
 = RC  time constant
At t = 0, (switch closed or open) a large current flows,
the capacitor behaves like a short circuit.
At t  , the current is essentially zero,
the capacitor behaves like an open switch.
The current decreases exponentially.
Measuring
Current and Voltage
in a circuit
The ammeter measures current,
and is connected in series.
The voltmeter measures voltage,
and is connected in parallel.
A modern digital multimeter combines the functions
of ammeter, voltmeter, and ohmmeter.
(i.e. can measure current, voltage, and resistance)
In addition, modern multimeters can measure
capacitance, temperature, and more,
and can be connected to computers too…