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Capacitors and Current Let’s start with a circuit consisting of a battery, a resistor, a capacitor and a switch in series: Start with no charge on the capacitor and the switch open. When the switch is closed at position c, current begins to flow, building up charge on the capacitor plates. At first, current flows as if there were a wire in place of the capacitor. As charge builds up on the capacitor plates, the voltage difference across the capacitor increases. This voltage difference across the capacitor opposes the battery voltage, decreasing the current flow. Once the voltage difference across the capacitor reaches the battery potential, current flow stops. So the increase in potential across the capacitor is proportional to the difference between the battery voltage and the potential across the capacitor. This proportionality between the rate of change of the potential and the difference between the potential and the battery voltage leads to an exponential curve. This curve representing the voltage across the capacitor must begin at zero, end at the voltage of the battery and have a decrease in the rate of change of voltage as time proceeds. Such a function is At t = , ΔV = 10 V. ΔV = 10 V [1-e-t/RC] With the capacitor fully charged, the switch is now closed to position d. The capacitor now loses charge driving current through the resistor until the plates are back to neutral. As the charge decreases, the voltage across the capacitor drops and the current through the resistor drops, decreasing the rate at which the voltage decreases. This gives another exponential curve. The charging and discharge curves for the circuit above are plotted below. Capacitor curves 1.20E+01 1.00E+01 voltage 8.00E+00 6.00E+00 4.00E+00 2.00E+00 0.00E+00 0 20 40 60 80 100 time (s) Notice the half time for the capacitor circuit is 5 seconds which is equal to RC ℓn(2). Note the units: Ohms Farads = seconds. Let’s check that. [Ω] = [V/A] = [V/(C/s)] = [Vs/C] [F]= [C/V] (from V = q/C) So, the units for RC are [Vs/C] [C/V] = [s]. So why doesn’t a break in the wire behave this way? A break in the wire doesn’t allow any noticeable current flow, just an open circuit. Let’s try a quick calculation using a 1mm diameter wire and a gap of 0.1mm between cut ends with air between. Treating this as a capacitor gives the following capacitance: C = o A/d = A/4Kd = (5E-4)2/[4(9E9)(1E-4)] ≈ 7E-14F Even with a 100kΩ resistor, this gives RC = 7E-9 seconds and very little charge on the “capacitor”. This means you would never notice the charge or discharge behavior of the open circuit, it’s over too quickly and little charge is transferred. RC circuits like this are often used for timing circuits. Logic circuits operate between 0 and +5 volts. A voltage between 0 and 2 volts is counted as a logical “zero” while a signal between 4 and 5 volts is counted as a logical “1”. Questions [1] If a 2μF capacitor is charged to +5 volts and you want it to drop to a logical 0 in 0.2 seconds. What resistor do you need to use to accomplish this? [2] Another use of RC circuits is in power filtering. To convert alternating current to direct current, diodes are used to give all positive voltages. The output from the diodes is shown in the diagram at right, the time between peaks is 1/120th of a second. The “ripple” needs to be minimized. The resulting voltage can be used to charge a capacitor and the discharge of the capacitor can be slowed using a resistor, shown in the following diagram. This signal results from the quick charging and slow discharge of the capacitor in the diagram. Typical capacitor values for such uses are 200 μF. If the resistor used is 330kΩ, how much does the voltage fall from 5.0 V. in the 1/120th of a second between peaks?