## Tuesday, September 10, 2019

### Charging and Discharging Capacitors Lab Report Example | Topics and Well Written Essays - 2000 words

Charging and Discharging Capacitors - Lab Report Example 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...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) = (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. 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). 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. While rising the current tends towards the maximum o