[SOUND] Welcome back to Linear Circuits. This is Doctor Weitnauer. This lesson is introduction to capacitors part one. Our objective is to understand the relationships between charge, current and voltage for capacitors. A cool application is supercapacitors or ultracapacitors. These are made by having two large area metal foils with molecularly thin separation and tightly packed, to make a large capacitance value in a small package. These are used to protect rechargeable batteries from current spikes. They are found in many applications, from small electronics to electric vehicles, and notice the size of the capacitance on these, the 10 farads and 500 farads. These are characteristic of supercapacitors or ultracapacitors. In contrast, traditional capacitors have much smaller values. For example, 0.3 micro farads. This lesson builds on the passive convention, which is the relationship between current's direction and voltage polarity, as shown. The charge and voltage on a capacitor have the relationship q=Cv. The capacitance C in Farads relates to the charge q in Coulombs and the voltage v in volts, as shown. The q represents the charge that is accumulated on one plate of the capacitor, the plate that is on the same side as the positive side of the polarity. And then in that case, on the other side of the capacitor, you'll have the same amount of charge but the negative of it. And then the v is the voltage across the device. In a plate capacitor, the capacitance C is equal to epsilon A over d, where A is the plate area, d is the spacing between the plates, and epsilon is the dielectric constant of the filler material. Capacitors are shown in schematics, like I show here. On the left, you see one of these has a curved bottom. That indicates it's a polarized capacitor, which must always have the positive polarity of the voltage on the flat side. Therefore, this type of capacitor is appropriate for DC applications only. Now, let's talk about current and voltage in a capacitor. First, remember, we're going to be assuming the passive convention. Current is the time rate of flow of charge. I equals dq/dt. So we differentiate q equals Cv to get dq/dt equals C dv/dt. Next, we substitute i in for dq/dt, giving us I equals c dv/dt. This is the iv characteristic for capacitors. If v is in Volts and C is in Farads, then i will be in Amps. It's important to note that, because the derivative operator is a linear operator, the i depends in a linear fashion on v, that makes the capacitor a linear element. So to sum up, we have the Q V characteristic: q = Cv and in the passive convention, we have the I V characteristic: i=C dv/dt. Thank you. [SOUND]