1• Capacitance and capacitorsECE 201: Lecture 15Borja PeleatoCapacitors• In its simplest form, a capacitor consists of two parallel metal plates separated by a dielectric barrier that does not allow the charges to go through• If a current pumps (positive) charges onto one plate, these charges will push the (positive) charges from the other plate away– The number of charges entering one plate is the same as the number of charges leaving the other, so we treat it as if the current were going THROUGH the capacitor (KCL holds)• In practice, the positive charges entering one plate accumulate there, while mostly negative charges remain on the other plate– This creates a voltage difference between the plates, which keeps increasing while charges keep arriving– If the current stops (no more charges arrive), the voltage remains constant. • Capacitors are represented by or 2Capacitance• The ratio between the amount of charge in the plates and the voltage that appears between them is known as Capacitance– Symbol: C– Units: Farads (F) = Coulombs/Volt. A capacitor of 1F would store 1C of charge on each plate (positive on one, negative on the other) when subject to a voltage of 1V– In general, the charge stored in a capacitor at time t is Q(t) = C*V(t)– The capacitance of a capacitor depends only on its shape and materials, not on the voltage or current it is subject to. For example, in a parallel plate capacitor3Voltage-Current equations• So far, we have studied the following laws– KCL, KVL: hold for everything (R,L,C)– Ohm’s law: holds for resistors only (V=R*I)– Inductor equations• Now we study the corresponding equations for a capacitor• Passive sign convention• Observe C-L parallelism: replace v(t) by i(t) and C by L 4Example• Now we are ready to analyze circuits:5Important properties• A capacitor can sustain a constant voltage with zero current, so for DC it behaves as an OPEN circuit (for inductors it was the other way around, they behaved like short circuits)• The integral equation shows that the VOLTAGE in a capacitor has to be CONTINUOUS, unless it gets an infinite current (for inductors, it was the current that had to be continuous)6Energy stored• Just like inductors, capacitors do not dissipate energy, they just store it. A charged capacitor will return all its charges to the circuit if needed• The power absorbed (or returned, depending on the signs and convention for v(t) and i(t)) is the same as always: P(t)=v(t)*i(t)• The energy stored in a given interval is– Does not depend on the waveform, just on initial and final voltages• The instantaneous stored energy is7Example89Example 2• Before the switch closes, the 1F capacitor is discharged and the 2F one has 6V across it. How much energy (in total) will the resistor dissipate after the switch closes?10Solution: Eventually, both capacitors will need to have the same voltage, lets call it Vf. Also, by conservation of charge, the total charge stored in both of them has to remain constant. Hence,2*6 = Vf*2 + Vf*1 so Vf = 4VBefore the switch closed, we had 0.5*2*6^2 = 36J stored in the capacitorOnce the voltages stabilize, we are left with 0.5*2*4^2+0.5*1*4^2 = 24J of energy stored between both capacitors.Consequently, the amount of energy that the resistor has dissipated is
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