To view the slide show with animations

1. To view the slide show with animations • Press F5 on the top of the keyboard • Security Alert – press OK (we have checked the photos) • Then to step forwards through the show, use the right arrow key • To go backwards, or repeat an animation, use the left arrow key • When you have had enough, press ESC key (top left of keyboard) CURRENT ELECTRICITY

2. w B y A C x z Circuit 1 Parallel Circuit 2 Series & parallel Demonstration 4Independent and Non-independent Branches What changes will you see in A, B C when RHS (w&x or y&z) is disconnected? CURRENT ELECTRICITY

3. Demonstration 4 circuit 1 OBSERVATION • A is unchanged when branch w&x is disconnected CONCLUSION • Parallel branches across battery are INDEPENDENT w n A A A x A o CURRENT ELECTRICITY

4. Demonstration 4 circuit 2 y&z branch disconnected OBS: brightness B decreases and C increases CONCL: B and C are not independent of y & z. B B y y z C C z Loops are NOT INDEPENDENT if they share a resistor (eg Bulb B) CURRENT ELECTRICITY

5. Demonstration 4 circuit 2 B and C have different current, so not in series y&z branch disconnected Brightness B > C > y = z Battery current is shared by C and yz C has less resistance than yz, so C more current When yz disconnected B (common to both loops) becomes dimmer – now yz has no current C becomes brighter – battery current not shared Now current B = C (series) B B y B and C now in series y z C C z CURRENT ELECTRICITY

6. A C B D TAKE NOTE: Series and parallel A and B in series B and F in parallel B and E in series A, E and F in series B in parallel to CD • False • False • False • True • True F E Current splits at node Geometry is not electrical parallel Current joins at node CURRENT ELECTRICITY

7. Demonstration 5: Kirchoff’s 1st Law • Predict the relationship between currents at A,B,C,D,E,F CURRENT ELECTRICITY

8. B A C E D F Demonstration 5 Compare the currents Node Node A = B = F (battery current) C = D (series current) B = C + E Conclusions Current can split or combine at a node (eg B or D) Current must split or combine at a node CURRENT ELECTRICITY

9. The total current going into a node equals the total current going out of a node n B x C A o y Circuit 2 Circuit 1 Kirchoff’s First Law (KI) node node CURRENT ELECTRICITY

10. Current relationships Node X • iA =iB=iF = battery current • iC = iD(series) • iB = iC + iE (at node x) • iD + iE = iF (at node Y) Kirchoff’s first law (K I)The total current going into a node equals the total current going out of a node Node Y CURRENT ELECTRICITY

11. p C D 6 A 8 A B q A Exercise 1 page 8 IA= 8 A (battery current) At node q: Iin= Iout(K I law) 6 A + IB= 8 A IB= 2 A IC = 2 A (series with B) ID = 6 A (given)        CURRENT ELECTRICITY

12. 2r i1 2.0 A r Current Splitting: Exercise 1Current is inversely proportional to resistance 2 Ω i2 6.0 A 3 Ω In these examples you are given one current and find another by inverse proportion CURRENT ELECTRICITY

13. 5r 0.5 A r i3 i4 2r For 3 or more resistors, compare them in pairs CURRENT ELECTRICITY

14. Exercise 2: Splitting up the TOTAL current r • For equal resistors • Current splits equally • i5= i6= ½ × 3.0 A • = 1.5 A i5 3.0 A i6 r CURRENT ELECTRICITY

15. For unequal resistors, current splits inversely I1 : I2 = R2 :R1 4  CHECK iTOTAL = 9.0 + 6.0 = 15 A I4 : I6= 6  : 4  = 6 : 4 (10 parts) i7 15 A i8 6  CURRENT ELECTRICITY

16. q p A D C B r s Exercise 3 (a)Identical bulbs 200 mA 0.100 A ID = 0.200 A (parallel to C) IA = IB = ½ × 0.200 A = 0.100A (A, B in series; 2 R gets ½ I ) At node s: K I law 0.200 A + 0.200 A= 0.400 A At node r: K I law 0.400 A + 0.100 A = 0.500 A Battery current = 0.500 A 0.200 A 0.200 A 0.500 A 0.100 A 0.400 A CURRENT ELECTRICITY

17. Exercise 3 (b) Identical bulbs i 120 mA C iA= ibattery (series) = 120 mA Share 120 mA in 2 branches iB: iCD = 2R : R = 2: 1 (3 parts inv prop) iB= = 80 mA iC= iD (series) = = 40 mA B D A CHECK : Itotal= 80 + 40 = 120 mA  CURRENT ELECTRICITY

18. 0.20 A 3R E R n m F R R G H q p 0.40 A J s r K Exercise 3 (c) 0.60 A 0.30 A iK = iJ (parallel, same bulbs) = 0.40 A iF= iR = × 0.20 A = 0.60 A iG = iH (series) = i2R = × 0.20 A = 0.30 A Using Kirchoff’s 1st law At n: Inq = 0.20 A + 0.60 A + 0.30 A = 1.10 A At s: Isq = 0.40 A+ 0.40 A = 0.80 A At q: Ibattery = 1.10 A + 0.80 A = 1.90 A 1.10 A 0.80 A 0.40 A CURRENT ELECTRICITY

19. Summary of current: General • Rate of flow of charge in a closed circuit • Measured in amperes (A). 1 A = 1 C s-1 • Inversely proportional to resistance of circuit • Battery current is not fixed; it depends on the components of the circuit and how they are connected. • The total current going into a node is equal to the total current coming out of that node (Kirchoff’s first law). CURRENT ELECTRICITY

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21. IA x Combined Series & Parallel RA I1 = IB RC I2 = IC = ID RB Ibattery RA and RB are NOT IN SERIES RB is parallel to RC& RD Ibattery= IA = IE (series) IC = ID (series) At x: IA = Ibranch1 + Ibranch2 (Kirchoff I) At y:Ibranch1 + Ibranch2 = IE (Kirchoff I) For parts given I1 : I2 = R2: R1 or For Whole Given Branch 1 Branch 2 RD IE y CURRENT ELECTRICITY

22. Well done. Now you need Part 2: Voltage To get back to the computer press ESC key (top left of keyboard) CURRENT ELECTRICITY