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RC Circuits 1. Draw a circuit diagram showing an RC circuit with a switch (as in figure 25.18 of your text). Assume the capacitor is initially uncharged. a) Before the switch is closed I= Vc = Vr = b) What is the voltage drop across the switch? At t= 0 s the switch is closed and current flows. 2. Write the loop equation for this circuit. 3. At the instant the switch is closed – you have the initial conditions for the circuit. Use the loop equation and knowledge of circuits to determine the initial value for: I= Vc = Vr = Qc = 4. Describe the changes in each of the initial values as current flows in the circuit – Qc Vc Vr I 5. After enough time has passed, the circuit reaches a final state (called steady state in engineering texts) where the values no longer change. What is the steady state value of each of the circuit parameters you have been evaluating? Solving the loop equation – be able to solve it at least one way 6. Rewrite the loop equation from part 1 as a first order differential equation. The general form will usually have all the terms with the variable on one side. You should be able to derive the solution to this equation by separating the variables and integrating. It is necessary to correctly identify the limits of integration. In this case the limits will go from the initial value of Q (which is 0 C for this example) the any final value, Q. Alternative Solution – evaluating the general solution Any differential equation of the form: (1) qQ/dt + aQ = C (constant) will have a solution for Q that is an exponential function. Since we can this as an exponential function it is know that the solution will have the form: (2) Q(t) = Ke-at + Qp Where K, a and Qp are all constants evaluated from the equation. This is a valid way of solving the equation that you should be able to perform. 7. Solving by evaluating the general form of the solution: a) Evaluate the particular (or steady state solution) – the steady state solution Qp is reached when there are no more changes in the system. So dQ/dt = 0. Solve for Qp when dQ/dt in equation (1) equals 0. b) Solve for the K using the initial conditions. Once you have a value for Qp, solve for K by solving for Q(t) = Q0 at t = 0 s. So in this case substitute Q0 = 0 at t = 0 and solve for K. c) Write the final form of the equation and check your answer using Ch 28 of the text. 8. Sketch a graph of Q vs t for this equation. Confirm that it exhibits the expected change in Q as the capacitor charges (explain) 9. Use the final solution for Q to find an expression for the current I as a function of time. Explain what this shows about current over time. 10. Sketch a graph of I vs.t. The Time Constant 11. The product RC has the units of time (go ahead and check). It is called the time constant. It is represented with the symbol τ. So τ= RC. When t = τ this is one time constant. a) Evaluation Q(t) and I(t) when t = τ. b) How would doubling R or C change the time constant? c) On your original sketch – add a line showing the Q vs t for a circuit with R (or C) doubled compared to the original. 12. Discharging the capacitorOnce the capacitor is charge, the battery can be removed without affecting the capacitor charge. When the loop is closed (figure 28.17) the capacitor discharges. a) Sketch the circuit and write the loop equation b) Solve this using either integration of solving the general form of: Q(t) = Ke-at Since the steady state in 0, you only have to evaluation the initial condition of Q =Q0 at t= 0 s ( but this time Q0 does not equal 0 C) Problems – refer to the examples in Chapter 28. 1. An uncharged capacitor and resistor are connected to a battery with EMF= 12 V, R = 4 Ω, C = 6µ F. The switch is closed at time t = 0. a) Find the time constant, the maximum charge on the capacitor and the maximum current in the circuit. b) Find the charge on the capacitor at t = 2τ c) Find the voltage across the capacitor at t = 2τ (Vc = 10 V) d) What is the voltage across the resistor at t = 2τ? 2. A capacitor of value C is discharging through a resistor, R. In terms of time constant τ = RC, when will the charge on the capacitor be one- half its initial value? (ans t = 0.69τ) b) When will the energy be one-half its initial value? (ans t=0.35τ)