[SOUND] Welcome back to Linear Circuits. This is Dr. Weitnauer. This lesson is Switching With Reactive Elements After DC. Our Objective is to learn what happens when a switch changes in a circuit with capacitors and inductors. That has been in DC steady state. It builds on the inductor current being continuous in time, capacitor voltage is also continuous in time. Inductor behaves as a short circuit to DC current. Capacitor behaves as an open circuit to DC voltage. We'll be talking about two switches. The first one is the single-pole single-throw or SPST switch. Here's a diagram of such a switch. It'll have an arrow that is labeled with a time. That means that the switch will either open or shut at that time. Now, the arrow indicates how it's going to change. In this case, the arrow is pointing up. Which means, that this switch is initially closed and then it opens at t = 0. That means that for negative time this is a short circuit and then, at t=0 becomes an open circuit. And here, I'm showing the arrow going the other way around, which means the switch is initially open and then closes at t = 0. The other kind of switch is the single-pole double-throw switch or SPDT switch. This is a three-terminal device and the connection is always made either between the a and b terminals or the c and a terminals. The way the arrow is showing here, it means that initially terminals a and b are connected. And then, to equal zero the switch changes and a and b become an open circuit and then a and c become a short circuit. This change is assumed to be instantaneous. The general approach to analyzing circuits like this type is, that you redraw the circuit prior to the switch change. You assume the circuit is at DC steady state or at rest. That means that all the derivatives are zero, inductors are short circuits, capacitors are open circuits. And then you calculate the inductor current or the capacitor voltage. Then you redraw the circuit after the switch change and you assume that those current or voltages that you calculated previously. They become the initial condition for the new circuit, that is, the new circuit after the switch change. Here's an example of an RC circuit with a single pulse, single throw switch. Before the switch changes state, assume that this circuit is at rest. and calculate v, the voltage across the capacitor at time t equals 0 plus, okay, 0 plus. The superscript plus means that it's the time immediately after the switch changes state. And then, we are also told that this circuit is in DC steady state conditions at that time. So that means, that not only are we going to draw the switch as a short circuit. But we're also drawing our compactor as an open circuit and the voltage that we are interested in is here. So, it's easy to see when the circuit is redrawn, that this is a voltage division problem. And the voltage divides evenly across the two resistors, so we will have that V is three volts. It's half the voltage and that is the voltage at time zero minus, which is right before the switch changes. And because of capacitors In capacitors, the voltage is continuous with time. It means that that is also the voltage immediately after the switch changes state. Then, if when it's time to determine the voltage for the remainder of time, for positive time, you would redraw the circuit again. This time, you don't include the middle branch, because that branch has been opened. And you have this RC circuit with an initial condition which is 3 volts. Prior to the switch change, we have a four resistor in the middle branch. We also have DC steady state conditions. Which means that the inductor is replaced by a short circuit, and we still have the other resistor, 4 amps. So if we find the current, then we see that that is a current division problem in this case because the two resistors are equal. The current divides the 3 amps, divides equally across the two branches and we have that i (0 minus). Which is just before the switch changes, is half of 3 amps or 3 half amps and because current in inductors is continuous in time. We also have that this is the same current through the inductor, just after the switch changes. After the switch changes, you have a new circuit. This time you have two ohms there. You have the inductor and you have four ohms over here. And what we know, in the lesson on first order RL students you would learn how to solve this. But you would need to know that i at 0, with 0+, is 3/2 amps for this second circuit. Okay, to sum up. Good practice is to redraw the circuit twice. Once before the switch changes and once after the switch changes. The inductor current or the capacitor voltage that you calculate before the switch change. Will become the initial condition for the circuit after the switch change. You will use those initial conditions when we do first order circuits. And you need the initial condition to get the solution to those circuits. Thank you. [SOUND]