RULES: Please try to work on your own. Discussion is permissible, but identical submissions are unacceptable! Please show all intermeate steps: a correct solution without an explanation will get zero cret. Please submit on time. NO late submission will be accepted. Please prepare your submission in English only. No Chinese submission will be accepted. 5.1 [10%] Suppose the two capacitors in Fig. 1 are initially uncharged. Then we connect a 15V source as shown. Find the voltages v 1 and v 2 at steady state by considering the conservation of charge on each plate. Fig. 1 Since both capacitors are initially uncharged, conservation demands that the net charge at each node remain 0 after the voltage source is connected. We apply this principle to the node between the two capacitors. Since we are looking at the negative plate of cap 1 and the positive plate of cap 2, we must have that q 1 q 2 = 0 [3pts] Remember that the charge on parallel-plate capacitors are equal and opposite. Another relation we can construct using KVL is the following: v 1 v 2 = q 1 C 1 q 2 C 2 = 15 [3pts] The above two equations give us a system that we can solve. The solutions are q 1 = q 2 = 0.126 mc [2pts]. With this we find that the voltages are v 1 = 10. 5 V [1pt] v 2 = 4. 5 V [1pt] 5.2 [8%] Determine L eq that may be used to represent the inductive network of Fig. 2 at the terminals. Fig. 2 1
Let the voltage across terminal a-b be v ab, and the input current be i. We know that v ab = L eq [2] (1) [2] By applying KVL, we can find that v ab = v 4H v 1 (2) [2] For the 3H and 5H inductor branches, v 1 = 2 v 2 (3) v 1 = 3 1 v 2 = 5 2 Solving equation (3) and (4), v 1 3 5 v 2 = 3 d(i 1i 2 ) v 1 = 21 8 = 3 (4) (5) Since v 2 = 5 8 v 4H = 4 (6) [2] Substituting (5) (6) into (2), v ab = 4 21 8 L eq = 53 8 = 53 8 = 6. 625 H 5.3 [10%] The current in a 4 H inductor is i = 10 A, t 0; i = (B 1 cos 4t B 2 sin 4t)e t 2 A, t 0. The voltage across the inductor (passive sign convention) is 60 V at t = 0. Calculate the power at the terminals of the inductor at t = 1s. State whether the inductor is absorbing or 2
delivering power. We know that p(t) = v(t)i(t), we can first get v(t): v(t) = 4 = 4 d (B 1 cos 4t B 2 sin 4t)e t/2 [2] v(t) = e t 2[( 16B 1 2B 2 ) sin 4t ( 2B 1 16B 2 ) cos 4t] V [2] Since i(0) = 10A, v(0) = 60V, i(0) = B 1 = 10A, v(0) = 2B 1 16B 2 = 60V B 2 = 5 [1] i(t) = (10 cos 4t 5 sin 4t)e t 2 A, t 0 v(t) = ( 170 sin 4t 60 cos 4t)e t 2 V, t 0 p(t) = ( 850sin 2 4t 600cos 2 4t 1400 sin 4t cos 4t)e t W [3] At t = 1s, p(1) = ( 850sin 2 4 600cos 2 4 1400 sin 4 cos 4)e 1 = 339. 57 W [1] Since p(1) < 0, the inductor is delivering power at t = 1s. [1] 5.4 [8%] a) Show that the two coupled coils in Fig. 3 can be replaced by a single coil having an inductance of L ab = L 1 L 2 2M. (Hint: Express v ab as a function of i ab.) b) Show that if the connections to the terminals of the coil labeled L 2 are reversed, L ab = L 1 L 2 2M. Fig. 3 a) b) a) v ab = L ab = L 1 1 L 2 2 M 2 M 1 = (L 1 L 2 2M) L ab = L 1 L 2 2M [4] b) v ab = L ab = L 1 1 L 2 2 M 2 M 1 = (L 1 L 2 2M) L ab = L 1 L 2 2M [4] 3
5.5 [10%] Consider the circuit in Fig. 4. Find i(t) for t < 0 and t > 0. Fig. 4 1. t < 0 Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. Apply nodal analysis for node A (let its node voltage be v A ), we can easily obtain 80 v A 0.5i = i = v A 40 80 Then we get v A = 64V and i = 0.8A [2] 2. t > 0 Let the current through the capacitor after the switch opened to be i 1. KCL: i 1 0.5i = i [2] For the capacitor, [2] Therefore, [4] Solving this equation, i 1 = C dv 1, v 1 = (30 50)i 240 = 0.5i 1 = 1. i 480 ln(i) = t 480 C 0, where C 0 is a constant number. t = 0, i = 0.8A, substituting this and we get C 0 = ln(0.8), ln ( i 0.8 ) = t 480 i(t) = 0. 8e t/480 A, t >0, or i(t) = 0. 8e t 480u(t) A, where u(t) is unit step signal. (Other correct methods are also accepted.) 4
5.6 [8%] For the op amp circuit of Fig. 5, let R 1 = 10 kω, R f = 20 kω, C = 20 μf, and v(0) = 1 V. Find v o. Fig. 5 For ideal op amp, we know that v 3 = 0. Then v 2 = v, and v 1 = 4u(t) At node 2, v 1 v 2 R 1 = C dv At node 3, 4u(t) v R 1 = C dv, v(0)=1v ln(4u(t) v) = t ln (3) R 1 C v(t) = 4u(t) 3e t/r 1C V [3] C dv = v 3 v o = v o R f R f v o = R f C dv = 3R f R 1 e t/r 1C [3] Substitute the values, or v o = 6e 5t V, t > 0 [2] v o = 6e 5t u(t) V 5.7 [10%] In the circuit shown in Fig. 6, the switch makes contact with position b just before breaking contact with position a. As already mentioned, this is known as a make-before-break switch and is designed so that the switch does not interrupt the current in an inductive circuit. The interval of time between making and breaking is assumed to be negligible. The switch has been in the position for a long time. At t = 0 the switch is thrown from position a to position b. a) Determine the initial current in the inductor. b) Determine the time constant of the circuit for t > 0. 5
c) Find i, v 1, and v 2 for t 0. d) What percentage of the initial energy stored in the inductor is ssipated in the 72 Ω resistor 15 ms after the switch is thrown from position a to position b? a) Before the switch is thrown from a, the circuit is in steady state and the inductor acts like a short circuit. Therefore, the initial current through it is I 0 =24/12 = 2 A. [2] b) The time constant is τ = L = 1.6 = 0. 02 s. [2] R eq 80 c) i(t) = I 0 e t/τ = 2e 50t u(t) A v 1 (t) = L = 160e 50t u(t) V Fig. 6 v 2 (t) = 72i(t) = 144e 50t u(t) V [3] d) At 0s, the initial energy stored in the inductor is w(0) = 1 2 Li(0)2 = 3.2 J At 15 s, the energy stored in the inductor is w(15) = 1 2 Li(15)2 = 0.714 J Therefore, the percentage of the initial energy ssipated in the 72 Ω resistor 15 ms is (3.2 0.714) 3.2 72 = 69.92% [3] 728 5.8 [8%] The switch in the circuit in Fig. 7 has been in position 1 for a long time. At t = 0, the switch moves instantaneously to position 2. Find v o (t) for t 0. Fig. 7 Initially the current through the inductor is 240 12440//10 4 5 = 8 A. [2] After the switch moved to position 2, the circuit can be equivalent to an inductor in series with 1 resistor, R eq = (46)//40 10 = 18 Ω. Let i(t) be the current through the inductor when t > 0. Then τ = L R eq = 0.004 s [2] 6
Since i(t) = I 0 e t/τ = 8e 250t u(t) A, we can find the current through the 40 Ω resistor to be Therefore, 10 1040 i(t) = 1.6e 250t u(t). [2] v o (t) = 1. 6e 250t u(t) 40 = 64e 250t u(t) V, t 0 [2] 5.9 [8%] The voltage across a 0.2 mf capacitor was 20V until a switch was opened at t = 0, causing the voltage to vary with time as v(t) = (60 40e 5t ) V for t > 0. a) Did the switch action result in an instantaneous change in v(t)? Why or why not? b) Did the switch action result in an instantaneous change in the current i(t)? Why or why not? c) How much energy was initially stored in the capacitor at t = 0? d) How much energy will be stored in the capacitor at t =? a) No, v 0 = 60 40 = v(0 ). The voltage v(t) across a capacitor is continuous. b) First we observe that i(0 ) = 0, since voltage is constant before the switch is thrown. After t = 0, we have i(t) = C = (0.2 mf)(200e 5t V) = 40e 5t ma Clearly, i(0 ) = 40 ma, so it is not continuous, leang to an instantaneous change. c) Energy is given by w(0) = 1 2 Cv2 (t = 0) = 1 2 (0.2 mf)(20v)2 = 40 mj d) After a long time, v(t) converges to 60 V. Using the same formula as before, we have w( ) = 0.36 J [2*4] 5.10 [10%] In Fig. 8, suppose that both switches have been open for a long time prior to t = 0. Then switch 1 closes at t = 0, followed by switch 2 at t = 10 s. Use MATLAB to plot v C (t) for t 0, assuming that v C (0) = 0. (Your MATLAB script should be attached!) At the time that the first switch closes, the capacitor sees the first 15 k resistor. Hence the fferential equation is v C 20 15k Fig. 8 (200μ) dv C = 0 [2] Rearranging the equation into a standard form gives us 7
v C 3 dv C = 20 Hence, we have a solution of the form v C = Ae t/3 B, where the first term is the homogeneous solution and the second is the particular solution. The time constant is simply the coefficient in front of dv C, so τ = RC = 3 s. The initial and final contions are v(0) = 0 and v( ) = 20, so the full solution is v C (t) = 20 20e t 3, 0 t 10 [3] When switch 2 closes, the equivalent resistance seen by the capacitor is now (15 k) (15 k) = 7.5 k. So the time constant is τ = RC = 1.5 s. The form of the solution remains the same, but we have fferent contions. The initial contion is equal to the voltage at t = 10 from before: v C (10) = 19.3 V. The final contion is v( ) = 10, as the capacitor is open in steady-state and we have a voltage vider. Hence our solution is v C (t) = 10 7308e t 1.5, t 10 [3] Graphically, we have a growing exponential to 10V, followed by a decaying exponential to 10V. [2] 5.11 [10%] Suppose the voltage source in the circuit of Fig. 9 is defined by a ramp function, such that v(t) = 0 for t < 0 and v(t) = t for t 0. If v C (0) = 0, derive an expression for v C (t) for t 0 and use MATLAB to sketch it to scale versus time. Consider trying a particular solution of the form v C (t)= A Bt. (Your MATLAB script should be attached!) Fig. 9 The fferential equation is v C (t) RC dv C = v(t) = t 8
If we use the standard homogeneous solution along with the suggested particular solution, we get v C (t) = A Bt De t/rc [2] The initial contion is that v C (0) = 0, so AD=0. Now if we plug in the solution into our original ODE: A Bt De t/rc RC (B D t e RC) = t [3] RC We now match terms to determine the coefficients' values. The only linear term on the LHS is Bt, so we must have B = 1. The constant terms are A RCB = A RC, and this must be equal to 0, since there are no constants on the RHS. Thus, A = -RC and D = -A = RC. The full solution is v C (t) = RC t RCe t RC = t RC (1 e t RC) [3] While we need a specific value for RC for an accurate plot, we can sketch a general characteristic by assigning a value of, say, 1 to RC. [2] Notice that the voltage originally exhibits a delay due to the exponential term. As time passes, the capacitor's voltage becomes linear and follows that of the source almost identically after it gets past the initial \inertia" presented by the capacitor. 9