Capacitors Capacitors store charge + ++ – – – E +Q –Q –––––––– d Q=CV Parallel plate capacitor plate area Dielectric permittivity ε=ε r ε 0 Units: farad [F]=[AsV -1 ] more usually μF, nF, pF 1

Capacitor types RangeMaximum voltage Ceramic1pF–1μF30kV Mica1pF–10nF500V (very stable) Plastic100pF–10μF1kV Electrolytic0.1μF–0.1F500V Parasitic~pF Polarised 2

Capacitors Capacitors in parallel: C1C1 C2C2 C3C3 Capacitors in series: C1C1 C2C2 C3C3 V1V1 V2V2 +Q 1 –Q 1 +Q 2 –Q 2 Q 1 =Q 2 3

Capacitors – energy stored Energy stored as electric field 4

RC circuits + V0V0 R C 1 2 Initially V R =0 V C =0 1 Switch in Position “1” for a long time. Then instantly flips at time t = 0. Capacitor has a derivative! How do we analyze this? There are some tricks… 1.) What does the Circuit do up until the switch flips? (Switch has been at pos. 1 for a VERY long time. easy) 2). What does the circuit do a VERY long time AFTER the switch flips? (easy) 3). What can we say about the INSTANT after switch flips? (easy if you know trick) I + + ALWAYS start by asking these three questions: 5

The Trick!! Note for Todd : Put problem on board and work through these points! 6

RC circuits + V0V0 R C 1 2 Initially at t = 0 - V R =0 V C =0 I=0 1 2 differentiate wrt t Then at t = 0 + Must be that V C =0 If V C =0 then KVL says V R =V 0 and I=V 0 /R. Capacitor is acting like a short-circuit. Finally at t = +∞ V R =0 V C =V 0 I=0 I 7

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+ V0V0 R 9

Integrator Capacitor as integrator R C ViVi V int If RC>>t V C <<V i 10

Differentiation R C ViVi V diff Small RC  11

Inductors I Solenoid N turns Electromagnetic induction B V ind 12

Self Inductance Mutual Inductance I I M Units: henrys [H] = [VsA -1 ] for solenoid 13

Inductors Wire wound coils- air core - ferrite core Wire loops Straight wire 14

Series / parallel circuits L1L1 L2L2 L3L3 Inductors in series: L Total =L 1 +L 2 +L 3 … L1L1 L2L2 Inductors in parallel I I1I1 I2I2 15

Inductors – energy stored Energy stored as magnetic field 16

The Next Trick!! 17

RL circuits + V0V0 R L R Initially t=0 - V R =V 0 V L =0 I=V 0 /R And at t=∞ (Pos. “1”) V R =0 V L =0 I=0 At t=0 +  Current in L MUST be the same I=V 0 /R So… V R =V 0 V R20 =20V 0 and V L = -21V 0 !!! I For times t > 0

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+ V0V0 R L Notice Difference in Scale! 20

+ V0V0 R2R2 R1R1 L I2I2 t<0 switch closed I 2 =? t=0 switch opened t>0 I 2 (t)=? voltage across R 1 =? Write this problem down and DO attempt it at home 21

Physics 3 June 2001 Notes for Todd: Try to do it cold and consult the backup slides if you run into trouble. 22

LCR circuit + V0V0 R C 1 2 L V(t) I(t)=0 for t<0 23

R, L, C circuits What if we have more than just one of each? What if we don’t have switches but have some other input instead? –Power from the wall socket comes as a sinusoid, wouldn’t it be good to solve such problems? Yes! There is a much better way! –Stay Tuned! 24

BACKUP SLIDES 25

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RL circuits + V0V0 R L 1 2 Initially V R =0 V L =

+ V0V0 R L 31

This way up 32

V0V0 R1R1 R1R1 R0R0 L R1R1 R0R0 L R1R1 R0R0 L R0R0 L Norton 33

This way up 34

This way up 35

Energy stored per unit volume Inductor Capacitor 36