1 EE 182 Midterm I [12:00] Name and ID: Last: First: Student ID #: Email: signature: 3/15/2022 Draw a box around the final answers otherwise you will NOT get any credits And move your solutions to the given boxes. Your phone must be turned off and kept in your bag. One (8.5×11) cheat sheet is allowed. Calculator is ok. No cell phone for calculator Only pencil and eraser 2 1. [8 7 15 pts] Assume that the op-amp is ideal amp. a) Find output voltage, ov b) Find the current flow over the 9kΩ with direction a) b) 9 5 Aki mΩ = ↑ a) 129v b) 5 Am ↑ 3 2. [3 pts each 15 pts] First order circuit is shown with DC input Vs and it is called CR circuit. The figure shows that the switch position is set (switch up) that way when 0t < and the switch will be flip-down at 0t ≥ a) Find voltage over the C and R when 0t < b) Find voltage over C and R when 0t ≥ c) Find current through C and R when 0t ≥ d) Plot voltage over C at [ ]0, 0,& 0t t t< = > e) Plot current over C at [ ]0, 0,& 0t t t< = > a) The voltage over the C is fully charged @ 0t < ( ) 12VCv t = And the current flow over the R is none since the capacitor is fully charged. So there is no current flow. ( ) 0VRv t = b) Only discharge occurs when 0t ≥ . It simply is a natural response a) ( ) VC sv t v= ( ) 0 VRv t = b) ( ) sC t v vt e τ = ( ) sR t v v et τ = c) ( ) sR t i v et R τ = ( ) sC t i v et R τ = d) e) 4 ( ) ( ) ( ) ( ) ( ) ( ) 0 0 1 t s t C R t t s c t s v t v t e v t i R where dC v t e R dt dC v e R dt C v e R C R d v e R C i t C v t dt τ τ τ τ τ τ τ + + = = = = = = = = = 1 sv RC t s t e v e τ τ = ( ) ( ) 1 1 t s R C t s t t s s v i t i t d R C v e dt C v e C v e e RC τ τ τ τ τ = = = = = 5 3. [5 pts each 20 pts] The switch position is set that way when 0t < (flipped down) and the switch position will be flip-up when 0t ≥ a) @ 0t ≥ , show the detail procedure that voltage across the capacitor and plot it. b) @ 0t ≥ , show the detail procedure that how you got the voltage over the resistor (no need to plot) c) @ 0t ≥ , show the detail procedure that how you got the current through the resistor and plot it. d) @ 10t ms= , how much power is delivered to the resistor a) ( ) ( ) ( )( )3 138.887.21012 1 12 1tC tv v et v e = = b) ( ) ( ) 37.2 1012C t v ei t = c) ( ) 37.2 100.005 t i t e = d) ( ) 3 7310 . wp ms mt = = a) The general equation for the DC input is ( ) ( ) ( )0 1t ty t y t e y eτ τ + ≥ = + ∞ The voltage over the capacitor is fully discharged 0t < since it is disconnected from the source, then ( ) ( ) ( )( ) 37.210 2 00 1 .4 3 .0072 12 1 t t e y t y e where RC k Fτ τ μ ≥ = ∞ = = Ω = = 6 b) The voltage over the resistor is ( ) ( ) ( ) ( ) ( ) ( ) ( )3 377.2 1 .0 2 103 1 13 12 2.47. 12 10 2 t t t Rv t i R dC v t R dt dC y e R dt F v e k v e τ μ = = = ∞ = Ω = c) The current through the resistor is controlled by this equation ( ) ( )di t C v t dt = ( ) ( ) ( ) ( ) ( ) 3 3 2 7.2 1 . 0 3 7 10 2 1 13 12 7.2 1 0.005 0.00 = 0 373w 3.73 t t tdi w e t C y e dt F v e p mi R τ μ = ∞ = = = = 7 d) The power delivered to the resistor at the given time is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 10 10 7.2 10 3 10 10 1 2 A .4 13 12 2.47.2 0.0012A 1.2m 10 t p t ms i t ms R dC y e k dt F v e k τ μ = = = = ∞ Ω = Ω = = 8 4. [15 pts] Plot voltage and power 9 10 5. [10 pts] a) Find ( )v t for the case of 0 ,0 &t + = ∞ b) Write ( )v t and plot it. c) Find ( )0.1v t ms= a) ( )0v ( )v ∞ τ ( )v t+ = b) c) +5V 11 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 2 20 1 2 : 5 2 2 4 5 2 2 : 5 2 2 3 : 4 5 2 2 : 5 2 5 6 6 5 2 5 5 3 1.11 @ . 3 0 0 277 1.1 1 .31 3 3 1 3 R x x x v L k i v i i i v i i L i i where i i i L i i v i i L i i i i v i t i t i < = Ω Ω Ω = Ω Ω = Ω + = Ω + = = Ω = + = = = = ( ) 1 1 2 1 2 3 3 1 2 3 2 1 2 3 2 3 1 3 1 2 3 2 2 3 1 2 3 3 1 3 1 2 1 1 : 0.277 : 4 2 5 5 : 5 2 5 5 : 4 2 5 5 2 : 5 2 5 5 5 5 5 2 21 1 5 25 5 sin x x L i L i i i i where L i i i L i i i i i i L i i i i i i i i i i Matrix is gular to working pr i i i i i i i = = + + = + + = + + + = = + + + = + = + + = + ecision 12 6. [7 6 2 pts] a) Find ( )v t for the case of 0 ,0 &t + = ∞ b) Write ( )v t and plot it. c) Find ( )0.1v t ms= a) ( )0 11 .754 V7v t = = ( )0 0.175vv t + = ( ) 0.5385vv t∞ = b) ( ) 0.3635 0.5385 0.01850 t y t e τ τ ≥ = + = c) ( ) 30.110 0.3635 0. .5385 70 1 0.1 7y vs et m τ += = = 13 a) ( ) ( )0 1@ 0 : 74 1.75t v t = = ( )@ 0t + 5.25V Using superposition property, 7 5.25 1 7 5.25 1 79 0.7 0.525 0. 17 0.52 1 75 0. 5 v v v v v v v v v v v v v vΩ = = + == + = = The value of ( )v t right after the flipping the switch is ( ) ( ) ( ) 0 3 0.17 1 5 0.5381 57 v t v vv t t v + = = ∞ = = ∞ = The equation for the voltage over the resistor is ( ) ( ) ( )( ) ( ) ( ) ( ) 0.3635 0.5385 0.0185 1 8 0.0183 50 0 || 0 8 0.175 0.53 5 0.5385 t t t y t y y e y e RC mF e τ τ τ τ τ + ≥ = + = ∞ + ∞ = = = = + = c) ( ) 30.110 0.3635 0. .5385 70 1 0.1 7y vs et m τ += = =
