Charging and Discharging Capacitors - Lab Report Example

www.academia-research.com Sumanta Sanyal d: 05/11/06 Transients: Charging and Discharging Capacitors Introduction This paper is on transient relationships in CR (capacitance-resistor) and LR (inductance-resistor) circuits. When a fully discharged capacitor is recharged the charge and voltage approach their maximum values exponentially according to' Kirchoff's Voltage Rule'. When a capacitor is being charged, at t=0, the voltage across the capacitor terminals is 0. As the charge builds up in the capacitor the voltage also increases. The increase is exponential, Kirchoff's Voltage Rule (Multi-loop Circuits, 1999), because as the charge builds up the rate of increase in voltage also increases. This is also true of the current through the circuit. Ultimately, at full charge, the voltage, ideally, becomes equal to that of the charging battery. In the case of LR circuits, the same is true of the current that increases exponentially according to Kirchoff's Current Rule. Transiently, when the switch is put on, the change in current is opposed by the back emf (rate of opposition decreases exponentially) until, at 5 time constant values, at 1 % accuracy, the change stops and the current reaches steady state. (Multi-loop Circuits, 1999) When charging the equations are differentiating and when discharging they are integrating. (University of Pennsylvania, 1998) Question 1: A capacitor of 12 microfarads capacitance is initially discharged. It is then charged through a 40 kilo-ohms resistor from a 50 volts d.c. supply for 0.2 seconds. Calculate: a) The capacitance voltage after the end of the 0.2 seconds On recharging, after some time 't': = (for an integrating circuit), Here, is the time constant which represents the minimum time the system needs to make significant change in voltage, charge and current. It is also called the 'resistor decay' and has the value 'RC' where 'R' is the resistance value in the circuit and 'C' is the capacitance value. Therefore, = RC = 40 x x F = .480 (time constant for the circuit) Therefore, = {1 - } = 50{1 - }(approx.) = 50{1 - .659} = 50{.341} = 17.05 V. (= 50V; Ideally, this battery voltage is achieved when the capacitor becomes fully charged.) b) The value of the charging current at the same instant: Since, I = = = 0.42625 amps = 426.25 mille-amperes c) The time constant value of the circuit. As already calculated in (a) time constant: = RC = 40 x x F = .480 Question 2: . A 20 microfarads capacitor is charged to 400 volts. At t = 0, a resistor of 12 kilo-ohms is connected across the capacitor. a) Calculate the voltage remaining across the capacitor after 180 milliseconds. In this case the capacitor is being discharged and the voltage drops from the maximum at t=0 to almost 0 when the discharge is complete. The drop is exponential as per Kirchoff's Voltage Rule. This time the voltage at time t is = = {}(for a differentiating circuit). The time constant for this circuit is: = RC = 12 x x F = .240 Therefore, voltage across the capacitor after 180 milliseconds: = {} = 400{} = 400{.472} = 188.8 V d) Calculate the discharge current at 180 milliseconds. Since, I = = = 0.0176 amps = 17.6 mille-amperes e) Calculate the time taken for the voltage to fall to 36.8% of its initial value. From the initial relationship: = {} it is derived: = .368, or = .368 t = = .24s The time taken by the voltage to fall to 36.8 % of its initial value is the time constant itself. Question 3: A coil having an inductance of 2.5 henrys and a resistance of 40 ohms is switched on to a 60 volt d.c. supply at t = 0. a) Calculate the value of the steady state current ultimately reached. The time constant in this case for inductance is: = = = .0625 Therefore, I = {1 - }= {1 - } = x .9933 = 1.49 amps b) Calculate the value of the current when t = 30 mille-seconds. Therefore, for the same circuit, I = {1 - } = {1 - }= .57 amps c) Determine the current value when t = seconds. Since is one time constant, I = {1 - } = {.632}= .948 amps Question 4: Fig. 1 (Appendix) At time t = 0 the switch is thrown to the left and after 5 seconds it is thrown to the right. The time constant = = = 1 a) Draw to scale waveforms to show the growth and decay of the current in L and the voltages across the two components R and L when the switch is operated to the left at t = 0 seconds and returned to the right after 5 seconds. For both integrating and differentiating equations the changing time component determines the gradient of the waveform and since steady state is reached after 5 time constants approximately, For growth and decay of current in L (Tabulated Values) Time t Rise in Current Fall in Current Current I {1 - } Current I 1 3.78 0.63 2.22 0.37 2 5.16 0.86 0.84 0.14 3 5.7 0.95 0.3 0.05 4 5.88 0.98 0.12 0.02 5 5.94 0.99 0.06 0.01 While rising the current tends towards the maximum of 6 amps but never reaches it actually and while falling the current tends towards the minimum of 0 but never reaches it. All values are approximations rounded off to two decimal places. (University of Pennsylvania, 1998) Waveforms 1 & 2 of the appendix depict the rise and fall in current respectively. Since the value of the circuit resistance is 1 ohms the rising and falling voltage values across the inductor are the same as those of the rising and falling current through it. Thus, no separate waveforms have been plotted as the voltage waveforms will be the same as the plotted waveforms for current through the inductance. For voltage across the resistance, the same situation prevails as the resistance value is 1 and the same waveforms computed for the rising and falling current will perfectly depict rising and falling current across the resistance. b) Calculate the time taken for the current to grow from 10% to 90% of its maximum value (the rise time) and the time taken for it to decay from 90% to 10% of its maximum value (the fall time). The value of steady state current at 1% accuracy: I = {1 - }, For this the time component {1 - } changes from 10% to 90%, = .9 (10%), and = .1 (90%); Taking natural logarithm, at 10% t = .1054s and, at 90% t = 2.3s Therefore, total rise time from 10% to 90% = 2.3 - .1054 = 2.195s (approx.) The same is true for the decay time: 2.195s (approx.) Conclusion All four questions relate to transient relationships and have been solved according to Kirchoff's Loop Rules. References All Exponential Values Calculated at: Bourne, David, 2006, Exponents. Accessed at (2nd, 3rd, and 4th November, 2006): http://www.boomer.org/c/p4/c02/c0202.html Multi-loop Circuits and Kirchoff's Laws, 1999. Extracted on 2nd November, 2006, from: http://physics.bu.edu/duffy/PY106/Kirchoff.html RC Circuits: Step Response and Exponential Waveforms, University of Pennsylvania, 1998. Extracted on 3rd November, 2006, from: http://www.seas.upenn.edu/ese205/Labs97S/Lab13.html/Lab13.html Appendix Figure 1: (For Problem 4) An LR circuit Waveform 1: Rise in Current with Time Waveform 2: Fall in Current with Time Read More