Voltage Calculation for Capacitor in Figure 1

1. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Problems With AssistanceModule 6 – Problem 2 Filename: PWA_Mod06_Prob02.ppt Go straight to the First Step Go straight to the Problem Statement Next slide

2. Overview of this Problem In this problem, we will use the following concepts: • Defining Equations for Capacitors Go straight to the First Step Go straight to the Problem Statement Next slide

3. Textbook Coverage The material for this problem is covered in your textbook in the following sections: • Circuits by Carlson: Sections #.# • Electric Circuits 6th Ed. by Nilsson and Riedel: Sections #.# • Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Section #.# • Fundamentals of Electric Circuits by Alexander and Sadiku: Sections #.# • Introduction to Electric Circuits 2nd Ed. by Dorf: Sections #-# Next slide

4. Coverage in this Module The material for this problem is covered in this module in the following presentation: • DPKC_Mod06_Part01 Next slide

5. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Problem Statement Next slide

6. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Solution – First Step – Where to Start? How should we start this problem? What is the first step? Next slide

7. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Problem Solution – First Step • How should we start this problem? What is the first step? • Write the defining equation for the capacitor, in integral form. • Write the equations for the lines in Figure 2. • Solve for the value of vC(0), the initial condition. • Find the value of iC(2[ms]). • Find the value of iC(8[ms]).

8. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for First Step –Write the defining equation for the capacitor, in integral form This is a good choice, and is the choice that we shall make. We are given the current through the capacitor, and we want the voltage. Thus, we want to use the defining equation for the capacitor in the integral form, since that makes it possible to find the voltage when we know the current. Let’s write this equation.

9. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for First Step –Write the equations for the lines in Figure 2 This is a reasonable choice for the first step, but is not the best choice. It is important to choose an optimal approach to a problem. In this problem, we can solve without needing these equations, using other methods to find the integral. In any case, the key is to start at the beginning. Then, we have the best chance to recognize the best next step. Please go back and try again.

10. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for First Step –Solve for the value of vC(0), the initial condition This can be done, but it is not a good choice for the first step. It is important to choose an optimal approach to a problem. In this problem, we can solve without finding the value of vC(0). In any case, the key is to start at the beginning. Then, we have the best chance to recognize the best next step. Therefore, we recommend that you go back and try again.

11. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for First Step –Find the value of iC(2[ms]) This can be done, but it is not a good choice for the first step. We are going to use this value as a part of our solution, but it is useful in most problems is to do things in a reasonable order. This will be something that we will need during a later step. It is better to start at the beginning. Therefore, we recommend that you go back and try again.

12. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for First Step –Find the value of iC(8[ms]) This can be done, but it is not a good choice for the first step. We are going to use this value as a part of our solution, but it is useful in most problems is to do things in a reasonable order. This will be something that we will need during a later step. It is better to start at the beginning. Therefore, we recommend that you go back and try again.

13. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Write the Defining Equation for the Capacitor Note: There must be the minus sign here, because the voltage and current are in the active sign convention. • We have defined the inductive voltage. What should the second step be? • Find the value of vC(0). • Set t0 = 0. • Set t0 = 2[ms]. • Set t0 = 2[ms] and t = 8[ms].

14. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Second Step –Find the value of vC(0) This is not a good choice for the second step. In this problem, we already have an initial condition given at 2[ms]. The initial condition does not always occur at t = 0, which is why we use the time t = t0 for the initial condition in the formula. Please go back and try again.

15. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Second Step –Set t0 = 0 This is not a good choice for the second step. In this problem, we already have an initial condition given at 2[ms]. The initial condition does not always occur at t = 0, which is why we use the time t = t0 for the initial condition in the formula. Please go back and try again.

16. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Second Step –Set t0 = 2[ms] This is not the best choice for the second step. This is a reasonable thing to do, since we have an initial condition given at 2[ms]. However, we can do more, given the nature of what we are looking for in this problem. Please go back and try again.

17. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Second Step –Set t0 = 2[ms] and t = 8[ms] This is the best choice for the second step. This is a reasonable thing to do, since we have an initial condition given at 2[ms]. In addition, we are looking for the value at 8[ms], so there is not reason not to go ahead and plug that time value in for t. This helps us to focus on what we need to do. Insert these values and go to next step.

18. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Inserting the Limits on the IntegralWhat is the Third Step? • We have inserted the limits on the integral. What should the third step be? • Find the equations of the straight lines in the plot. • Find the total area under the curve for all time (t). • Find the area under the curve for the time period 2[ms] < t < 8[ms].

19. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Third Step –Find the equations of the straight lines in the plot This is not the best choice for the third step. This is a reasonable thing to do, if we plan to perform the integral. This would be the general approach, and will certainly give the correct answer. However, in this case we can find the integral more easily by finding the area under the curve. Please go back and try again.

20. The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Your Choice for Third Step –Find the total area under the curve for all time (t) This is not the best choice for the third step. We can certainly find this area, but it is not what we are looking for in this problem. Rather, we are looking for the integral from 2[ms] to 8[ms]. Thus, this total area is not useful. Please go back and try again.

21. Your Choice for Third Step –Find the area under the curve for 2[ms] < t < 8[ms] The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). This is the best choice for the third step. We are looking for the integral from 2[ms] to 8[ms]. Thus, the area in this time range is what we need. We can find this area, shaded in the plot above, by taking the difference between the areas of triangles. Let’s find this area.

22. Finding the Area Under the Curve for 2[ms] < t < 8[ms] The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). This area for each shaded section can be found with the difference of the areas of two triangles. Now, we can plug this back in to the equation.

23. Plugging in, to get the Solution The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is given that vC(2[ms]) = 6[V]. Find vC(8[ms]). Plugging this value into the equation, we get Go to the Comments Slide

24. These Units are Weird. Do I Have to Show All These Units? • Of course, the answer to any question like this should be given by your instructor, so we will not try to answer this question here. • However, units are important. Many people handle units in this kind of problem by converting all the units to “base units”, that is, [V], [A], [W], [F], [H], [W], [J], [C], [s], and so forth. Then, it will be clear that any results that are obtained will come out with base units. In this case, the units are usually not shown until you reach your answer. • In any case, be careful with your units! Go back to Overviewslide.