Problem Set 5 Solutions


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1 University of California, Berkeley Spring 01 EE /0 Prof. A. Niknejad Problem Set 5 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different ways. 1. (a) No. v(t) is continuous, as v(0) = (60 0e 0 ) V = 0 V = v(0 ). (b) First we observe that i(0 ), since voltage is constant before the switch is thrown. After t, we have i(t) = C dv(t) = (0. mf)(00e 5t V) = 0e 5t ma Clearly, i(0 ) = 0 ma, so it is not continuous, leading to an instantaneous change. (c) Energy is given by w(0) = 1 Cv (t ) = 1 (0. mf)(0 V) = 0 mj. (d) After a long time, v(t) converges to 60 V. Using the same formula as before, we have w( ) = 1 Cv (t = ) = 1 (0. mf)(60 V).36 J.. Under DC conditions, all capacitors become open circuits. We redraw the circuit, along with some nodal voltages: v 3 0 kω v x v y 30 kω 5 kω v 1 v 15 kω 15 V v kω Since no current flows across the two leftmost resistors, we have v 1 = v = v x. Also, v 3 = v x v 7 and v = v y. To find v x and v y, we can use a simple voltage divider: v x = 15 (15 V) = 1.5 V 5 15 v y = (15 V) = 5 V 5 15 So v 1 = v = 1.5 V, v 3 = 7.5 V, v = 5 V.
2 3. The required energy is E = P t = (1.5 kw)(0 60s) = 3.6 MJ. So we need a capacitance given by C = E (3.6 MJ) = = 180 F. The amount of charge is V (00 V) Q = CV = (180 F)(00 V) = 36 kc. As a parallel plate capacitor, it would have an area of A = Cd 180 ɛ rɛ 0 = 6 = 1.36 (15)( ) 6 m.. At the time that the first switch closes, the capacitor sees the first 15 kω resistor. Hence the differential equation is v C 0 15 k (00 µ) dv C Rearranging the equation into a standard form gives us v C 3 dv C = 0 Hence, we have a solution of the form v C (t) = 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 conditions are v(0) and v( ) = 0, so the full solution is v C (t) = 0 0e t/3, 0 t When switch closes, the equivalent resistance seen by the capacitor is now (15 kω) (15 kω) = 7.5 kω. So the time constant is τ = R eq C = 1.5 s. The form of the solution remains the same, but we have different conditions. The initial condition is equal to the voltage at t = from before: v C () = 0 0e /3 = 19.3 V. The final condition is v C ( ) =, as the capacitor is open in steadystate and we have a voltage divider. Hence our solution is v C (t) = 7308e t/1.5, t Graphically, we have a growing exponential to V, followed by a decaying exponential to V v C (t) t
3 5. At the time of the switch action, both capacitors have 0 V across them, as they act as open circuits. They do not change instantaneously, so they remain at 0 V each after the switch opens. Thus, the voltage drop across the resistor remains 0, and no current flows. We would also expect the current to be 0 at steady state at t =, so i(t) for all time. You would get this solution even if you were to solve the governing ODE due to the 0 boundary conditions. 6. The differential equation is v C (t) RC dv C(t) = v(t) = t If we use the standard homogeneous solution along with the suggested particular solution, we get v C (t) = A Bt De t/rc The initial condition is that v C (0), so A D. Now if we plug in the solution into our original ODE: A Bt De t/rc RC (B DRC ) e t/rc = t 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 ) 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 v C (t) t 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. 3
4 7. (a) Writing a KCL equation at v(t) gives us v C (t) 500 k ( µ)dv C(t) = i(t) = ( µ)e t v C (t) dv C(t) = 5e t (b) The time constant is τ = RC = 1 s, the coefficient in front of the derivative term. The complementary (homogeneous) solution thus takes the form v 0 (t) = Ae t. (c) If we were to use a particular solution of the form v p (t) = Ke t, we get the same form as that of the complementary solution. But the complementary solution only works for a homogenous ODE (i.e., the forcing function is 0): v C (t) dv C(t) = Ke t Ke t (d) If we were to try v p (t) = Kte t instead and plug it into the ODE, we now obtain Kte t Ke t Kte t = 5e t So the constant is K = 5. Combining it with the homogeneous solution, we have v C (t) = (A 5t)e t. If we assume that the capacitor is initially uncharged, then v C (0) and A. So our full solution is v C (t) = 5te t. 8. In using superposition, we consider one source at a time. Let s deal with v(t) first. Then the current source becomes an open circuit, and the equivalent resistance seen by the capacitor is 1 kω (the two 1 k are in series, and they are in parallel with the k). Hence the time constant is τ = R eq C = (1 kω)(0.5 µf).5 ms. The forcing function (voltage source) is constant, so we seek a solution of the form v O1 (t) = A Be 000t. Using the conditions v O1 (0) and v O1 ( ).5 V (the capacitor becomes an open circuit, and we have a simple voltage divider at v O1 (t)), the full solution is v O1 (t).5(1 e 000t ). Next we have the current source, and v(t) becomes a short. To make this easier to deal with, we can use a source transformation to turn the current source and resistor into a Thévenin equivalent with a voltage source (alternatively, you can write some KCL equations). 1 kω 1 kω v O (t) kω 1 V 0.5 µf Notice that the circuit is nearly identical to the previous one. So the solution is (without resolving it) v O (t).5(1 e 000t ). The full solution is the sum of the two individual components: v O (t) = 1 e 000t.
5 9. The golden rules hold, as the op amp is in negative feedback. Defining the node between the capacitor and resistor in the feedback loop to be v x, we can write KCL equations at v x and at the inverting input. C d(v x 0) v x v o R 0 v i C d(0 v x) R 1 We can integrate the second equation to solve for v x and plug it back into the first: v x (t) = 1 v i (t) v i(t) 1 v i (t) v o(t) R 1 C R 1 R 1 R C R v o (t) = R v i (t) 1 v i (t) R 1 R 1 C. Since the op amp merely serves as a buffer, we have v i (t) on the left side of the capacitor. The voltage on the right side is i o (t)r by Ohm s law. Then KCL gives us i o (t) C d(i o(t)r v i (t)) i o (t) RC di o(t) = C dv i(t) We cannot simplify further since we do not know the form of v i (t). We can infer that i o (t) will have a homogeneous solution Ae t/rc, but the form of the particular solution depends on v i (t). 5
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