The charge Q stored by a capacitor is given by Q = CV .
So basically, what happens in the DC circuit, the volt ampere characteristics of a capacitor in the stage of the steady state are in where the capacitors prevent the flow of DC current, but it will indeed charge up to a DC voltage. Clearly speaking, the basic characteristics of a capacitor in the steady state stage are the ones who are not short circuit, but an open circuit. The charge stored by the capacitor is identified in the form of: Q = CV. The energy stored in the capacitor is determined by the form: W = ½ CV^ 2 . The figure below illustrates that the capacitor will be charging up to the amount of voltage expressed in (E) in five-time constants, this is in where we derive the equation to be: τ = RC In one time constant, the voltage Vc will charge up to 63.2% of its final value, in 2τ up to 86.5%, in 3τ up to 95.1%, in 4τ up to 98.1%, and in 5τ up to 99.3%. That now brings us to a conclusion in where Vc = E (1-e^- t/ RC ) and i c = E/R (e^- t/ RC). Now let us consider the following circuit: The measured resistance from what we have obtained in the multimeter was 11.183 kΩ. Next, we are going to figure out the steady state value of the current (I) and voltages (V 1 and V 2 ). τ 0 1 2 3 4 5 V 2 0 6.3212 8.64 6 9.50 2 9.811 9.9326
For the two voltages: V 1 = Ee^- τ = 10 e ^-5 = 0.06737 V V 2 = Vmax (1-e^-t/ τ) To note down: By 5 the value reaches the max. For the current: I = V/R = 10 / 1.2 x 10^ 3 = 8.33 mA Yet again we will have to compute the voltages (V1 and V2) and the current (I) by using ohm’s law. Given to us V= 6 mV V=IR I = V/R = 6 x 10^-3 / 1.2 x 10^3 = 4.62 µA V 1 = 0.06739 V V 2 = 9.9326 V Compared to results we obtained in the first part we can clearly witness that the only difference was with the current, in where we clearly got a different answer. Now we will be calculating the energy stored by the capacitor using the equation: W = ½ CV^ 2 W= ½ (100 x 10^-6)(100)^ 2 = 5 mJ We will then disconnect the supply and measure the voltage across the disconnected capacitor. Now what basically happened when we were testing it, it discharged at 0.6 s. The capacitor went from full charge to 0 in 0.6 s. One final thing that had to be considered was that we had to short circuit terminals with a lead and then do the measuring all over again. This step was necessary because it helped identify if there is any flow of charges about the circuit without the capacitor being there.
Section 3 (Parallel R-C DC Network) The main aim in this part of the experiment is to get a better clearance about parallel R-C networks.
- Spring '20
- Resistor, Electrical impedance, Cramer