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Capacitor problems 1. A 470 F capacitor is charged by a 20V cell and then discharged across a 530 resistor. a) Draw a suitable circuit to achieve this. b) Show that the time constant for the discharge circuit is 0.25 s. c) Show that the units of the time constant are seconds (s). d) Show that the p.d. across the capacitor has fallen to 7.4 V after 0.25 s. e) Plot a graph to show the p.d. across the capacitor after 0.25, 0.50 and 0.75 s. Answers: a) Suitable circuit to include cell, capacitor, 2-way switch, Voltmeter across the capacitor. b) Time constant = RC = 530 x 470 x 10-6 = 0.249 s. c) Use Q/Qo = e-t/RC to show RC is in s. d) After one “RC”, p.d. has dropped by e: 20/e = 20/2.7 = 7.4V e) After 0.50 s, V= 7.4/e = 2.7 V. After 0.75 s,V = 2.7/e = 1.0 V. Plot suitable graph. 2. Use the circuit below for this question: Cell: 6.0V A B Capacitor: 4700 F Resistor: 5.6 k a) When the switch is moved from A to B, what will happen to the capacitor current? b) Show the initial value of the current is about 1mA. c) What is the time constant of the circuit? d) What is the current flowing in the circuit after the switch has been closed for a time equivalent to 2 x RC. Answers: a) The current will move in the opposite direction to the charging current, and through the resistor as the capacitor DISCHARGES. b) Since I =V/R, I is 6/5600 = 1.1 x 10-3 A c) Time constant = RC = 4700 x 10-6 x 5.6 x 103 = 26.3 s. d) After 2 x RC seconds, I will be 1.1mA/e2 = 1.1/7.29 = 0.15 mA. 3. A capacitor is used to make a light flash on and off at regular intervals. The capacitor is marked 4700 F and is attached to a 100 V cell and a resistor. a) Find the charge on the capacitor. b) If the time constant of the circuit is 0.70s, what is the value of the resistor in the circuit? c) When the capacitor charges and reaches 72V, the lamp switches on. What is the energy stored in the capacitor? d) When the lamp flashes, it transfers energy at an average rate of 150 W.What is the duration of the flash? e) Draw a suitable circuit to show how the lamp can be switched on and off. Include all components. Answers: a) Q = CV = 4700 x 10-6 x 100 = 0.47 C b) 0.70 = 4700 x 10-6 x R. So R = 150 c) E = ½ CV2 = ½ x 4700 x 10-6 x 72 x 72 = 12.2 J d) t = 12.2/150 = 0.08 s e) Suitable circuit includes capacitor and lamp in parallel with R, switch and cell in series.