1. INC 111 Basic Circuit Analysis Week 9 RC Circuits

2. 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 V 0 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 RL Source-free RC

5. t v(t) V0V0 or t i(t) V 0 /R

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 I 0 is the initial current of the circuit

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

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

11. Response time Period 1 How to Solve Problems? Period 2 Period 3 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.

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 condition of 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. t i(t) 4A 2A

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 condition of 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. t i(t) 1.66mA t vc(t) 5V 3.33V

21. Example Find and sketch i C (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 0.8 A ic(t)