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CAPACITORS © John Parkinson 1 BASIC CONSTRUCTION The Parallel Plate Capacitor CONDUCTOR + - INSULATOR © John Parkinson CONDUCTOR TWO OPPOSITELY CHARGED CONDUCTORS SEPARATED BY AN INSULATOR WHICH MAY BE AIR 2 CAPACITORS STORE • CHARGE & • ENERGY USES 1. Storing energy as in flash photography 2. Time delays in electronic circuits 3. As filters in electronic circuits 4. In tuning circuits © John Parkinson 3 Charge stored [Q] depends on p.d. [Volts] applied [V] Q Gradient = C = V Q V Hence Q = C V C is measured in FARADS, [F], - more often : F, nF or pF The capacitance, C, is defined as the charge required to raise the potential by one volt. © John Parkinson 4 Charging a Capacitor Q Q + + + - C Q Q = CV +Q -Q © John Parkinson 5 As VOLTS = Joules per Coulomb WORK DONE BY THE CELL W=QV The pd across the capacitor builds up as more charge is added Voltage V Charge 0 Q 0 V 1 W [ ] Q QV 2 2 © John Parkinson Work done in charging = average volts x charge 6 The work done in adding such an infinitesimally small amount of charge, q, that V remains constant is Charge Q given by: Volts V v q w = V.q This is the area of the strip. Hence the total work done = the energy stored in the capacitor is the area under the graph 1 E QV 2 Or if we combine with Q = CV, or © John Parkinson 1 E CV 2 2 1 Q2 E 2C 7 What is the charge and the energy stored when a 50 F capacitor is charged to (a) 100 V, (b) 200 V? (a) Q = CV = 50 x 10-6 x 100 = 0.005 C E = 0.5CV2 = 0.5 x 50 x 10-6 x 1002 = 0.25 J (b) Q = CV = 50 x 10-6 x 200 = 0.01 C E = 0.5CV2 = 0.5 x 50 x 10-6 x 2002 = 1.00 J Why does doubling the p.d. in (b) quadruple the energy stored?? © John Parkinson 8 Charging a Capacitor Initially Vc = 0, so V = Vr and so initial current = ? V VR VC R C At all times after the switch is closed V = VR + VC V I0 R Finally I =0 when capacitor is charged and VC = V Volts I0 VC time © John Parkinson 0 time 9 CAPACITOR DISCHARGE Initially V = V0 + - V I0 R Hence initial current, I0 = ? V R I Final Current = ? Volts, V V0 V V0e 1 t RC i.e. the decay is exponential What if R and / or C is larger? 0 © John Parkinson Time, t 10 CAPACITOR DISCHARGE V V0e 1 t RC Q Q0 e The current also falls exponentially and is given by : 1 t RC I I 0e 1 t RC Current I I0 Note the area under this graph is the initial charge stored. Time, t © John Parkinson 11 CAPACITOR DISCHARGE R times C V V0e has units of 1 t RC seconds time constant and is called the of the circuit If we put t=RC into the equation above, it becomes V=V0 e-1, which works out to: V=0.37V0 i.e. when time is equal to R x C the p.d. across the capacitor has dropped to 37% of its original p.d. The capacitor is almost discharged in 5 time constants. Try putting t = 5 x R x C © John Parkinson 12 CAPACITOR DISCHARGE V V0e 1 t RC Taking logs to the base “e” 1 ln V ln V0 t RC In the form y = c + m x Hence Plot ln V Gradient is negative and equal to 1 RC t © John Parkinson 13 The Parallel Plate Capacitor Area of Plate overlap = A Medium relative permittivity = r d A 0 r C d © John Parkinson d = plate separation 0 = the permittivity of free space = 8.86. X 10-12 F m-1 For air or a vacuum, r = 1 14 Question 1 A 1000 F capacitor is charged to 100 V and then discharged through a 2000 resistor. (a) What will be the initial current? (b) What will be the current after 4 s? (a) (b) V 100 I 0.050 A R 2000 V V0e t RC 100e 4 20001000106 100e2 13.5V V 13.5 I 0.00675 A 6.75 mA R 2000 © John Parkinson 15 Question 2 A capacitor is charged to 100 V and then discharged through a resistor whose value is gradually reduced in order to maintain a constant current of 200 mA though it. The capacitor becomes discharged after 10 s. (a) What is the initial value of the resistor? (b) What is the capacitance of the capacitor? (a) (b) Initially the p.d. is 100 V, so V 100 R 500 I 0.200 Q = I t = 0.200 x 10 = 2.0 C The capacitor was initially charged to 200 V with 2.0 C of charge Q = C V, so 2.0 = C x 100 Hence C = 2 x 10-2 F or 20 000 F © John Parkinson 16