1 / 26

INC 111 Basic Circuit Analysis

INC 111 Basic Circuit Analysis. Week 9 RC Circuits. RC Circuits. The response of RC circuits can be categorized into two parts: Transient Response Forced Response Transient response comes from the dynamic of R,C. Forced response comes from the voltage source. Source-Free RC Circuits.

leason
Download Presentation

INC 111 Basic Circuit Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. INC 111 Basic Circuit Analysis Week 9 RC Circuits

  2. RC Circuits The response of RC circuits can be categorized into two parts: • Transient Response • Forced Response Transient response comes from the dynamic of R,C. Forced response comes from the voltage source.

  3. Source-Free RC Circuits Capacitor has some energy stored so that The initial voltage at t=0 is V0 Initial condition Find i(t) from R, C

  4. Compare with the solution of RL circuits. The solution of RC circuits can be obtained with the same method. Source-free RC Source-free RL

  5. v(t) V0 t i(t) V0/R or t

  6. Time Constant The product RC is time constant for RC circuits Unit: second

  7. Forced RC Circuits C has an initial voltage of 0 from Use KVL, we got

  8. From Differentiate both sides Solve first-order differential equation Where I0 is the initial current of the circuit

  9. Natural Response Force Response C has an initial voltage = 0, But from KVL, therefore,and So,

  10. vC(t) i(t) V/R V t t vR(t) Note: Capacitor’s voltage cannot abruptly change V t

  11. How to Solve Problems? • Divide in to several periods (3 periods as shown below) • Period 1, 3 have constant V, I -> Use DC circuit analysis • Period 2 is transient. Response Period 3 Period 1 Period 2 time

  12. Calculate Transient (period 2) • Start by finding the voltage of the capacitor first • Assume the response that we want to find is in form of • Find the time constant τ (may use Thevenin’s) • Solve for k1, k2 using initial conditions and status at the stable point • From the voltage, find other values that the problem ask using KCL, KVL

  13. Example Switch open for a long time before t=0, find and sketch i(t) First, we start by finding vc(t) The initial condition of C is vc(0) = 1V The stable conditionof C is vc(∞) = 3V

  14. Assume vc(t) in form of Find the time constant after t=0 by Thevenin’s, viewing C as a load Therefore, the time constant is

  15. Find k1, k2 using vc(0) = 1, vc(∞) = 3 At t=0, vc(0) = 1 V At t = ∞, vc(∞) = 3 V Therefore, k1=3, k2 = -2 We can find i(t) by using Ohm’s law on the resistor

  16. i(t) 4A 2A t

  17. Example The switch was opened for a long time before t=0, Find i(t) Start with vc(t) The initial condition of C is vc(0) = 5V The final stable conditionof C comes from voltage divider, which is vc(∞) = 5*(1/1+0.5) = 3.33V

  18. Assume vc(t) in form of Find the time constant after t=0 by Thevenin’s, viewing C as a load Therefore, the time constant is

  19. Find k1, k2 using vc(0) = 5, vc(∞) = 3.33 At t=0, vc(0) = 5 V At t = ∞, vc(∞) = 3.33 V Therefore, k1=3.33, k2 = 1.66 We can find i(t) by using Ohm’s law on the resistor

  20. i(t) 1.66mA t vc(t) 5V 3.33V t

  21. Example Find and sketch iC(t)

  22. Assume vc(t) in form of Use Thevenin’s, viewing C as a load to find Req Therefore, The time constant is

  23. Find initial condition vc(0)

  24. Find initial condition vc(∞)

  25. Find k1, k2 using vc(0) = 1, vc(∞) = 5 At t=0, vc(0) = 1V At t = ∞, vc(∞) = 5 V Therefore, k1=5, k2 = -4 We can find ic(t) by

  26. We found that ic(t) = 0 for t < 0 ic(0-) = 0 ic (0+) = 0.8 ic(t) 0.8 A

More Related