Download Real Batteries and Capacitance Notes UPDATE

Real vs. Ideal Batteries
Capacitance
Battery vs Cell
• Cell: something that converts chemical energy
into electrical energy, usually via an
electrolyte and two substances with different
conductivities.
• Battery: combination of two or more “cells”
E
Why a battery works…
Electric Potential Energy: the - end has a greater potential
energy than the + end so there is a change in potential
across the cell, or potential difference (voltage) of the cell.
• As current flows through the cell itself, every unit of
charge undergoes an increase in potential energy. This
increase in potential energy is given by:
E=Vq
(E=energy in J, V=voltage, q=charge in C)
Therefore, the voltage of a cell is equal to the energy
provided by the cell per unit of charge passing through it.
Ideal Battery
Ideal Battery: can supply an unlimited current, but
the flow is limited to a fixed and finite value by the
circuit (resistance).
-Terminal voltage is static (Vs)
Real Batteries
Real Batteries: have an internal resistance that is equivalent to an
ideal cell in series with a resistor.
ℰ
r
Electromotive force (EMF), ℰ (symbol): simply the energy
provided by the ideal cell per coulomb of charge passing through it:
ℰ=E/q
• E is the energy provided by the cell to the charge, and Q is the
amount of charge that passes through the cell
• The EMF can also be defined as the potential difference across
the cell when NO current is flowing . The only time the EMF
value is equivalent to the terminal voltage.
• Not a force
Real Batteries
Internal resistance (r): the resistance of the
real cell
• fairly constant in value for most cells if they
are not mistreated.
• a battery with internal resistance cannot be
an ideal battery as its own resistance will
limit the current
– also affects the current in the circuit in which it
is connected
Real Batteries in a Circuit Diagram
Real Battery
ℰ
r
IB
R
Circuit diagram showing a cell (two plates of a battery)
connected to a resistor of resistance R. The cell has fixed
internal resistance r and an EMF “ℰ” as shown. A current of
magnitude IB flows through the circuit.
Calculating Voltage in a Real Battery
In the previous circuit, the EMF and current are related as follows:
ℰ =I(R+r)
where I is the current flowing through the circuit, “R” is the resistance of
the load in the circuit and “r” is the internal resistance of the cell. This is
found simply by using V=IR.
This can be rearranged to give:
and then:
ℰ =IR+Ir
ℰ =VB+Ir
• where VB is the terminal potential difference (the potential difference
across the terminals of the real cell when current is flowing through
the circuit).
This can be further rearranged to make VB the subject, such that:
VB= ℰ −Ir
Real Batteries
Voltage vs. Current Graph
ℰ
-r
VB= ℰ −Ir
• ℰ is voltage with zero current flowing in the circuit.
– Resistance must be infinite
• Voltage drops as current increases, slope is -r
• Io is maximum current with zero resistance in the circuit, voltage is
now zero (no pressure with zero resistance)
Real Battery Calculations
Question 1: A circuit is set up identical to the
one above. If R=230Ω, ℰ=12.0V and I=0.05A,
find the internal resistance, r.
Solution: Vc = V1 + V2
ℰ =IR + Ir , solve for r
ℰ −IR
𝒓=
I
r= 10Ω
Battery Combinations
Series: adding batteries in series results in adding
voltage of each battery together.
VB=2 V
VB=2 V
VB=2 V
(Resistor)
VB=2 V
Battery Combinations
Parallel: batteries in parallel are equivalent to their
original battery voltage
VB=2 V
VB=2 V
VB=2 V
VB=2 V
• If identical batteries are used, then each battery will also provide the
same current
• If non-identical batteries are used, a few things can happen. The voltage
of the cells will be balanced, the difference in cell properties such as
internal resistance will affect how the current goes in and out of each cell,
or if one cell is shorted, it can drain the other cell.
Battery Combinations
• Real Batteries
– Follow previous rules for multiple batteries to find
voltages
– Internal Resistance
• Use rules for resistances to find equivalent resistances of
battery combinations
For example, two 2V batteries added in series each with an
internal resistant “r” equal to 1 Ω would be handled as
such:
• Total Voltage= 2V+2V=4V
• Total Internal Resistance= 1 Ω + 1 Ω = 2 Ω
For parallel configurations, follow appropriate parallel rules.
Capacitance
• Capacitance: the ability to store charge
• Capacitor: an electric circuit component
capable of storing charge
– Symbol: C
– Unit: Farad (1 Coulomb/1 Volt)
• Common Capacitors:
– parallel-plate capacitor: two conductive plates
insulated from each other, usually sandwiching a
dielectric (poor electrical conductor) material.
• capacitance is directly proportional to the surface area of
the conductor plates and inversely proportional to the
separation distance between the plates
Capacitance Equations
𝑪=
𝒒
∆𝑽
C = Capacitance (Farads)
q= charge (coulombs)
V= voltage (Volts)
𝑪=
𝑨
𝒌𝜺𝟎
𝒅
k=relative permeability for type of
medium (1 for vacuum)
𝜀0 = 8.854x10-12 Farad/meter
(constant for permittivity of vacuum)
A=area of one plate (m2)
d=distance between plates (m)
Capacitance Equations Explained
𝑪=
𝒒
∆𝑽
This solves for charge
on a capacitor
𝑪=
𝑨
𝒌𝜺𝟎
𝒅
This determines size
of the capacitor
• Permittivity: measure of resistance that is
encountered when forming an electric field in a
medium.
• The dielectric lowers the electric field strength
for charges on a capacitor which lowers the
voltage for the same charge and therefore
increases capacitance.
Capacitance Equations Explained
q /∆𝑽 = 𝑪 = 𝒌
Constant
2) Efield
decreases
causing
Voltage to
decrease
𝑨
𝜺𝟎
𝒅
Constant
1) Dielectric
increases
3) Capacitance
increases
Capacitance Equations… One More
1 2
𝐸 = 𝐶𝑉
2
• Finds the energy stored in a capacitor
• Remember that W = ΔE
Sample Calculation
Question 2: Find the capacitance of a capacitor if 5 C’s
of charge are flowing and potential applied is 2 V.
Solution:
Given: Charge Q = 5 C, Voltage V = 2 V
Equation: C = Q/V
= (5C)/(2V)
= 2.5 F. (F is Farads)
Capacitance Calculation with Area and
Distance
Question 3: Calculate the capacitance of a capacitor having
dimensions, 30 cm X 40 cm and separated with a distance of
8mm of air gap.
Solution: A=0.30m x 0.40 m = 0.12 m2
k = 1 for air
𝜀0 = 8.85 x 10 -12 C2/N.m2 (constant)
d= .008 m
𝐶=
𝐴
𝑘𝜀0
𝑑
= (1) (8.85 x 10
C=1.3275.10-10 F
-12 C
2
.12 𝑚2
2
/N.m )(
.008 𝑚
)
Capacitance in a Circuit
• Drawing capacitors in a circuit:
– Two Symbols
• Two Parallel Plate = basic
• One Curved, One flat plate: capacitor that can only be
used with DC circuits because it has polarity
+
-
Capacitors in a Circuit Diagram
Series Circuits
Parallel Circuits
Calculating Equivalent Capacitance in a Parallel Circuit
Parallel Rules and Derivation
Using rules for current in a parallel circuit:
IT= I1 + I2 + IN , substitute I =
𝑞𝑡
𝑡
=
𝐶∆𝑉𝑡
𝑡
𝑞1
𝑡
+
𝑞2
𝑡
=
𝐶∆𝑉1
𝑡
+
+
𝑞𝑁
𝑡
𝐶∆𝑉2
𝑡
𝑞
∆𝑡
for each
Came from :
𝐶=
𝑞
∆𝑉
, then substitute each for 𝑞 = 𝐶∆𝑉
+
𝐶∆𝑉𝑁
𝑡
, time and voltage would cancel
Times all the same, voltages all the same (VB= V1 =V2 =VN parallel)
Leaving
C T = C1 + C 2 + CN
Calculating Equivalent Capacitance in a Series Circuit
Series Rules and Derivation
Using rules for voltage in a series circuit:
VT = V1 + V2 + VN
𝑄𝑡
𝐶
=
𝑄1
𝐶
+
𝑄2
𝐶
+
𝑄𝑁
𝐶
, using 𝑄 = 𝐶∆𝑉 𝑟𝑒𝑎𝑟𝑟𝑎𝑟𝑎𝑛𝑔𝑒𝑑
since Q is current per time and both are constant values and
would cancel ( IT= I1 + I2 + IN)
Leaving
1
𝐶𝑇
=
1
𝐶1
+
1
𝐶2
+
1
𝐶𝑁
Capacitance in a Circuit Calculations
Question 4: Find capacitance if capacitors of 6 F and 5 F
are connected
(i) In series
(ii) In parallel.
(i) Solution: Equation:
1
𝐶𝑇
=
1
𝐶1
+
1
𝐶2
=
1
𝐶𝑇
=
1
6
+
1
5
= 2.73 F.
(ii) Solution: Equation: CT = C1 + C2 = 6 + 5 = 11 F
Practice
• Do Capacitors Problem Set 1
• Do appropriate text problems on capacitance