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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.

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INC 111 Basic Circuit Analysis

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  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

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