Physics 106 Lesson #16
Capacitors
Dr. Andrew Tomasch
2405 Randall Lab
atomasch@umich.edu
Review: Ohm’s Law
• The resistance of the
resistor (light bulb) is R
• The voltage across the
resistor is V
• The current through the
resistor is I
• Ohm’s Law:
V  IR
Ohms (W)
I
Voltage = V
R
Voltage = 0
I
Assume perfect
wire  NO voltage
drop across wires!
I
Review: Adding
Series Resistors
• For resistors R1 & R2
connected in series
(sequentially), the
current i passing
through each resistor
must be the same
• The voltages across
R1 & R2 must add up
to V → V = iR1 + iR2
= i(R1 + R2) = iReq
V
+
Req  R1  R2
Review: Adding Parallel Resistors
• Resistors R1 & R2
connected in parallel
have the same potential
difference (voltage) V
across them
• The total current is the
sum of the current
through each resistor
• I = I1 + I2 = V/R1 + V/R2 =
V(1/R1 + 1/R2 ) = V /Req
V
+
1/ Req  1/ R1  1/ R2
Review: More Ways to Calculate Power
• Electric Potential
Energy is transformed
into to some other
form (heat, light) by the
resistor (light bulb).
• Power:
V  IR
I V /R
I
Voltage = V
R
Voltage = 0
I
2 Electric Potential
V
P  IV  I R 
R
2
(Energy) decreases
across the light bulb
(resistor)
I
Review: Superconductivity
Zero Resistance
• The resistance of many (but
not all) substances
decreases with temperature
• A special class of materials
known as superconductors
lose all electrical resistance
below a temperature known
as the critical temperature
• Superconductors can
levitate magnets by the
Meissner Effect where a
perfect conductor expels all
magnetic fields from its
interior
Capacitance
• Most objects
can be charged
and therefore
store charge.
• Capacitance is
charge stored
per electric
potential: Unit:
Q
C
V
Coulombs/Volt = Farads
(for Michael Faraday)
Capacitance of a Parallel-Plate Capacitor
Q
C
V
A
C 
d
For a parallel-plate
capacitor, the
capacitance is
proportional only
to geometric
factors (the area of
the plates A and
the separation
distance between
them d) and the
natural constant 
 C is fixed at the
factory!
Dielectrics
• Capacitance is
increased by adding a
dielectric (increase ε):
C
A
d
• More charge is stored
per volt with larger ε
compared to air
   0
Vacuum
Energy Stored in a Capacitor
Units: Joules
2
1
1
Q
2
Energy  QV  CV 
2
2
2C