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=energy in J, V=voltage, q=charge in C) 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. 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 ℰ r

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 IB r 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: ℰ =IR+Ir and then: 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 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) ℰ -r VB= ℰ −Ir

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 𝒓= ℰ −IR 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 VB=2 V (Resistor)

Battery Combinations Parallel: batteries in parallel are equivalent to their original battery voltage 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. VB=2 V VB=2 V VB=2 V VB=2 V

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=𝒌 𝜺 𝟎 𝑨 𝒅 C = Capacitance (Farads) q= charge (coulombs) V= voltage (Volts) k=relative permeability for type of medium (1 for vacuum) 𝜀 0 = 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 𝑪=𝒌 𝜺 𝟎 𝑨 𝒅 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. This solves for charge on a capacitor This determines size of the capacitor

Capacitance Equations Explained Constant Constant 2) Efield decreases causing Voltage to decrease 1) Dielectric increases 3) Capacitance increases

Capacitance Equations… One More 𝐸= 1 2 𝐶 𝑉 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 𝜀 0 =8.85 x 10 -12 C2/N.m2 (constant) d= .008 m 𝐶=𝑘 𝜀 0 𝐴 𝑑 = (1) (8.85 x 10 -12 C2/N.m2)( .12 𝑚 2 .008 𝑚 ) C=1.3275.10-10 F

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 𝑞 𝑡 𝑡 = 𝑞 1 𝑡 + 𝑞 2 𝑡 + 𝑞 𝑁 𝑡 , then substitute each for 𝑞=𝐶∆𝑉 𝐶∆𝑉 𝑡 𝑡 = 𝐶∆𝑉 1 𝑡 + 𝐶∆𝑉 2 𝑡 + 𝐶∆𝑉 𝑁 𝑡 , time and voltage would cancel Times all the same, voltages all the same (VB= V1 =V2 =VN parallel) Leaving CT = C1 + C2 + CN , substitute I = 𝑞 ∆𝑡 for each Came from : 𝐶= 𝑞 ∆𝑉

Calculating Equivalent Capacitance in a Series Circuit Series Rules and Derivation Using rules for voltage in a series circuit: VT = V1 + V2 + VN 𝑄 𝑡 𝐶 = 𝑄 1 𝐶 + 𝑄 2 𝐶 + 𝑄 𝑁 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 𝐶 𝑇 = 1 𝐶 1 + 1 𝐶 2 + 1 𝐶 𝑁

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 𝐶 𝑇 = 1 𝐶 1 + 1 𝐶 2 = 1 𝐶 𝑇 = 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