Physics 169. Luis anchordoqui. Kitt Peak National Observatory. Wednesday, March 8, 17

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1 Physics 169 Kitt Peak National Observatory Luis anchordoqui 1

2 5.1 Ohm s Law and Resistance ELECTRIC CURRENT is defined as flow of electric charge through a cross-sectional area Convention i = dq dt Unit Ampere [A = C/s] Direction of current is direction of flow of positive charge Current is NOT a vector but the current density is a vector ~j = charge flow per unit time per unit area i = Z ~j d ~ A 2

3 Drift Velocity ˆ Consider a current i flowing through a cross-sectional area A ) In time q n t total charges passing through segment Q = qa(v d t)n charge of current carrier density of charge carrier per unit volume ) Current i = Q t = nqav d Current Density ~j = nq~v d 3

4 Note For metals charge carriers are free electrons inside ) ~j = ne~v d for metals ) ~j ~v d Inside metals and are in opposite direction We define a general property of materials conductivity ( ) Note In general Resistivity ( ~j = ~ E is NOT a constant number but rather a function of position and applied = 1 ) is more commonly used property defined as m Unit of : Ohm-meter ( ) where Ohm ( ) = Volt/Ampere ~E -field OHM S LAW Ohmic materials have resistivity that are independent of applied electric field e.g. metals (in not too high ~E -field) 4

5 Example Consider a resistor (ohmic material) of length ) L and cross-sectional area Electric field inside conductor V b V a = Z b a ~E d~s = E A Z L 0 dx = EL ) E = V L Current density j = i A ) = E j Note R V i = V = ir is NOT a statement of Ohm s Law V L = R = L A 1 i/a resistance of conductor but it s just a definition for resistance 5

6 ENERGY IN CURRENT Assuming a charge Q enters with potential V 1 and leaves with potential V 2 ) ) Potential energy lost in wire Rate of energy lost per unit time Joule s heating For a resistor R U = QV 2 QV 1 U = Q(V 2 V 1 ) U = Q t t (V 2 V 1 ) P = i V = Power dissipated in conductor R, P = i 2 R = V 2 R 6

7 5.2 DC Circuits A battery is a device that supplies electrical energy to maintain a current in a circuit In moving from point 1 to 2 electric potential energy increase by U = Q(V 2 V 1 )= Work done by E Define E = Work done/charge = V 2 V 1 7

8 Example o V a = V c assuming perfect conducting wires V b = V d By Definition V c V a V d = ir V b = E ) E = ir ) i = E R Also we have assumed zero resistance inside battery 8

9 Resistance in combination Potential difference V a V b =(V a V c )+(V c V b ) ) Equivalent Resistance = ir 1 + ir 2 R = R 1 + R 2 1 R = 1 R R 2 for resistors in series for resistors in parallel 9

10 Example For real battery there is an internal resistance that we cannot ignore ) E = i(r + r) E i = R + r Joule s heating in resistor P = i (P.D. across resistor R) Question What is value of Answer We want to find Setting R R R dp dr = E 2 E 2 2R (R + r) 2 (R + r) 3 dp dr =0) E 2 = i 2 R E 2 R P = (R + r) 2 to obtain maximum Joule s heating? that maximizes [(R + r) 2R] =0 (R + r) 3 ) r R =0 ) R = r P P max max P r 2r 3r R 10

11 ANALYSIS OF COMPLEX CIRCUITS KIRCHOFF S LAWS First Law (Junction Rule): Total current entering a junction equal to total current leaving junction Second Law (Loop Rule): The sum of potential differences around a complete circuit loop is zero Convention (i) (ii) V a >V b ) = ir Potential difference i.e. Potential drops across resistors V b >V a ) Potential difference =+E i.e. Potential rises across negative plate of battery 11

12 Example By junction rule: By loop rule: i 1 = i 2 + i 3 (6.1) Loop A ) 2E 0 i 1 R i 2 R + E 0 i 1 R =0 Loop B ) i 3 R E 0 i 3 R E 0 + i 2 R =0 (6.2) (6.3) Loop C ) 2E 0 i 1 R i 3 R E 0 i 3 R i 1 R =0 (6.4) BUT (6.4) = (6.2) + (6.3) 12

13 General rule Need only 3 equations for 3 current i 1 = i 2 + i 3 (6.1) 3E 0 2i 1 R i 2 R =0 2E 0 + i 2 R 2i 3 R =0 (6.2) (6.3) Substitute (6.1) into (6.2) 3E 0 2(i 2 + i 3 )R i 2 R =0 (6.4) ) 3E 0 3i 2 R 2i 3 R =0 Subtract (6.3) from (6.4), i.e. (6.4)-(6.3) 3E 0 ( 2E 0 ) 3i 2 R i 2 R =0 ) i 2 = 5 4 E0 R 13

14 Substitute i 2 into (6.3) 2E E0 R ) i 3 = Substitute i 2,i 3 into (6.1) R 2i 3 R =0 3 8 E0 R i 1 = E0 8 R = 7 8 E0 R Note A negative current means that it is flowing in opposite direction from one assumed 14

15 5.3 RC Circuits (A) Charging a capacitor with battery Using loop rule Note Direction of i +E 0 {z} ir Q {z} C =0 P.D acrossr P.D across C is chosen so that current represents rate at which charge on capacitor is increasing ) E = R i z} { dq dt + Q C ) dq EC Q = dt RC 1st order differential eq. 15

16 Integrate both sides and use initial condition Z Q 0 dq EC Q = t =0, Q on capacitor =0 Z t 0 dt RC ln(ec Q) Q = t t 0 RC 0 ) ln(ec Q) +ln(ec) = t ) ln 1 ) Q EC Q EC = e t/rc = t RC ) Q EC =1 e t/rc ) Q(t) =EC (1 e t/rc ) RC 16

17 Note At As Current t =0,Q(t = 0) = EC (1 1) = 0 t!1,q(t!1)=ec (1 i = dq dt = EC n i(t = 0) 1 RC i(t) = E R e t/rc i(t!1) = 0 e t/rc 0) = EC E R = Initial current = i 0 17

18 At time =0 capacitor acts like short circuit when there is zero charge on capacitor!1 As time open circuit. capacitor is fully charged and current =0 it acts like a open circuit c = RC is called time constant It is time it takes for charge to reach q I 1 1 Q 0 ' 0.63 Q 0 e C 0.632C ε ε I 0 ε I 0 = R τ τ =RC t 0.368I 0 τ t 18

19 (B) Discharging a charged capacitor Note Direction of i is chosen so that current represents rate at which charge on capacitor is decreasing Loop Rule ) i = dq dt V c ir =0 ) Q C + dq dt R =0 ) dq dt = 1 RC Q 19

20 Integrate both sides and use initial condition Z Q Q 0 dq Q = 1 RC Z t 0 dt ) ln Q ln Q 0 = Q t ) ln = Q 0 RC ) Q Q 0 = e t/rc t 0, Q on capacitor = Q 0 t RC i = dq dt ) Q(t) =Q 0 e t/rc ) i(t) = Q 0 RC e t/rc ( t = 0) ) i(t = 0) = 1 At R i 0 = V 0 R Q 0 {z} C Initial P.D. across capacitor 20

21 At t = RC = Q(t = RC) = 1 e Q 0 ' 0.37 Q 0 21

22 5.4 Semiconductors A semiconductor is a substance that can conduct electricity under some conditions but not others making it a good medium for control of electrical current Specific properties of a semiconductor depend on impurities added to it (or dopants) N-type semiconductor carries current in form of negatively-charged in a manner similar to conduction of current in a wire e P-type semiconductor carries current predominantly as e deficiencies called holes A hole has a positive electric charge equal and opposite to charge on e In a semiconductor material flow of holes occurs in a direction opposite to flow of electrons 22

23 Acronym LED stands for Light Emitting Diode A diode is a combination of two semiconductors one of which is doped n-type and other p-type (region between n-type and p-type material) When electrons and holes meet in junction they can recombine to liberate energy 23

24 5.5 Intuition Each of the 12 edges of a cube contain a 1 Ω resistor Calculate the equivalent resistance between two opposing corners 24

25 Here is where the intuition comes into play Color coding is used to help keep track of the resistors and associated nodes Due to symmetry P.D. at the three nodes labeled " " are equal Since no current flows between nodes with a potential difference of zero V, they can be shorted together without affecting the circuit's integrity The same can be done for the nodes labeled " " Once you short those nodes you obtain the following equivalent circuit 25

26 There are two sets of three resistors in parallel in series with one set of six resistors in parallel So you have 1/3 Ω in series with 1/6 Ω in series with 1/3 Ω which equals 5/6 Ω 26

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Chapters 24/25: Current, Circuits & Ohm s law Thursday September 29 th Register your iclickers

Chapters 24/25: Current, Circuits & Ohm s law Thursday September 29 th **Register your iclickers** Conductors under dynamic conditions Current, current density, drift velocity Ohm s law Types of conductor