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t W I w v 6.00-fall 017 Midterm 1 Name Problem 3 (15 pts). F the circuit below, assume that all equivalent parameters are to be found to the left of port oa A. Find the Thevenin equivalent resistance TH.5 / B. Find the Thevenin equivalent voltage V; TH Sh o.t A t VV I ovt

t T N V S 6.00-Fall 017 Midterm 1 Name a C. Find the Norton equivalent resistance Circuit reproduced for convenience T# + #/:I D. Find the Norton equivalent current / VrN A I,o +

Problem 5 Part (A) When the motor does not turn, Ω = 0, and using the matrix relation for the transformer, the transformer voltage from the electrical side, v, is also 0 since v = K e Ω = 0. Therefore, all of the voltage V S is dropped across the 50Ω resistor, the motor coil resistance, and the coil inductance L. Furthermore, at steady state, the current through L is not changing and the the voltage drop across it, v L = L di dt = 0. Therefore, at steady state V S is dropped completely across 50Ω and. Since 50Ω and are in series, we can use voltage divider to get: V scope,steadystate = earranging the above equation yields + 50 V S = 83.4mV = 50Ω 16.6 mv = 9.95 Ω 100mV = 16.6 mv + 50 From the plot of V scope we can see that it takes 50µs to decay from 100 mv down to 47.8 mv. This decrease of 5.7 mv in 50µs corresponds to 63% of the difference between the intial and final (steady state) value of V scope. Therefore, the time constant of this first-order decay is τ = 50µs. Furthermore, the equivalent resistance as seen from the terminals of the inductor L is T h = + 50Ω. We can use the L circuit charging/discharging time constant formula to calculate the inductance, L, of the coil: τ = 50 10 6 = L τ = 50 10 6 = T h L + 50Ω Substitute = 9.95 Ω, as calculated above L =.998 mh Problem 5 Part (B) The plot shown is for the decay of electrical current I. Therefore, we need to compute the effective capacitance, C eq, of the mechanical moment of inertia J as seen from the electrical ports, i.e., the mechanical capacitance J after it is reflected through the transformer. On the mechanical side, we know that as the applied torque, T, flows into the moment of inertia, the angular velocity Ω across it increases, resulting in an equation similar to that of the electrical capacitor, but with mechanical variables, T and Ω: T = J dω dt We can use the matrix equation for the transformer to substitute T and Ω in the above equation with their equivalent electrical variables of v and i (which is the same as I because of the series circuit). From the given transformer matrix equation, we know v = K e Ω and i = T/K e. These substitutions result in the following equation: ik e = J d( v K e ) dt i = J dv Ke dt 1

dv This can be compared to the constitutive relation for a capacitor, i.e., i = C eq dt, thus, showing that the mechanical moment of inertia behaves like an electrical capacitor of equivalent capacitance C eq = J/Ke. This equivalent capacitance is in series with and the voltage source V s. Therefore, the time constant, τ I for the decay of current I, as shown in the plot on Page 13, is given by: τ I = C eq = 10 10 3 s J = τ I K e τ I = J K e = 0.01s 4(V/(rad/s)) 9.95Ω J = 4.0 10 3 kg m Problem 5 Part (C) We showed in Part (B) that the mechanical moment of inertia, J, looks like a capacitor of capacitance J/K e, when reflected through the transformer to the electrical domain. Similarly, mechanical damping B r, represented in the mechanical domain as a resistor of resistance 1/B r, has the following constitutive relation in the mechanical domain: Ω = 1 B r ( T ) From the given matrix equation, we know v = K e Ω and i = T/K e. These substitutions result in the following equation: v K e = 1 B r ik e v = K e B r i This can be compared to the constitutive relation for a resistor, i.e., v = eq i, thus, showing that mechanical damping behaves like an electrical resistor of equivalent resistance eq = K e /B r. In the electrical domain to the left of the transformer, this equivalent resistance of mechanical damping is in parallel with C eq of the moment of inertia because they share the same angular velocity Ω across their terminals on the mechanical side of the transformer and that Ω is transformed to the same voltage v across both of their terminals on the electrical side. Therefore, after reflecting the mechanical components through the transformer, the resulting circuit looks like: + i + V s + V scope v K e B r J K e

Problem 5 Part (D) The resulting system is a first-order C circuit with a time constant given by τ C = T h C eq, where T h is the equivalent Thevenin resistance seen from the J/Ke capacitor port. To compute T h, we set the voltage source V S to zero and replace it with a short (or a wire). Then, the equivalent resistance from the capacitor port is just the parallel combination of and K e B r : K e B r T h = K e = B r + K e B r = K e B r + K e Using this T h and C eq = J/K e, we can compute the time constant: K e τ C = T h C eq = B r + Ke J τ C = B r + Ke J K e Problem 5 Part (E) The equivalent resistance that contributes to the observed time constant is a parallel combination of two resistances, and K e /B r. Without any additional information or measurements, we can only extract the effective resistance from the time constant, and are unable to further make a distinction between the two contributing resistances. Problem 5 Part (F) In the case where the electrical port is open-circuited, i = 0. Therefore, no current flows into the transformer from the electrical side. From the given transformer matrix equation, we know that: i = T K e = 0 T = 0 [N m] Therefore, no torque flows into the transducer as long as the electrical port is open-circuited. As such, for the given circuit, any externally applied torque will flow only into the moment of inertia branch and the mechanical damping branch. Problem 5 Part (G) In a manner analogous to electrical circuits, the moment of inertia of the motor acts like a short during the impulse of torque at t = 0, which causes all of the applied torque to flow into the moment of inertia branch and deposit all of the the angular momentum, T 0, on the motor s moment of inertia. At t = 0+, the torque impulse has passed and the independent torque source can be replaced with an open-circuit. The angular momentum deposited by the impulse of torque causes the motor to start rotating with some angular velocity, Ω(0+), which slowly decreases as energy is dissipated due to damping losses a discharging process which is represented by the torque flowing out from the moment of inertia capacitor and through the damping resistor, resulting in a decreasing angular velocity Ω(t) of the motor, until the motor stops rotating completely. During the impulse: T 0 δ(t) = J dω dt 3

T 0 J T 0 J T 0 δ(t)dt = dω J 0+ 0 0+ 0 δ(t)dt = 0+ 0 dω δ(t)dt = Ω(0+) Ω(0 ) From the plot of Ω vs. time on Page 16, we know Ω(0+) = rad/s and Ω(0 ) = 0. Also, from definition, we know 0+ 0 δ(t)dt = 1. Substituting these values into the equation above, we get: T 0 J J = T 0 J = T 0 1 = rad/s [ ] Nms rad/s kg m Additionally, we know that the time constant for the motor s angular velocity decay after the impulse is 1 second, and it is given by: τ mech = mech C mech = 1 B r J = J B r = 1second J B r = 1second B r = T [ ] 0 kg m or s [ ] Nms rad 4