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HW4_Soln

HW4_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 4...

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 4 Solutions Capacitors and Resistor Circuits (plus Diodes) Page 1 of 12 1. [8 pts total] Analyze the circuit at right and find the voltages at nodes A and B. 1 1 2 3 2 1 2 3 1 2 4 KVL 1: (2V) (10k )I (2k )I (1k )I 0 KVL 2: (5k )I (1k )I (1V) 0 KCL 1: I I I 0 KCL 2: I I I 0 ! " ! " + " = " ! " ! = ! ! ! = + + = From KCL 1 & 2 1 2 3 I I I + = ! & 1 2 4 I I I + = ! I 3 = I 4 We only need three equations! 1 2 2 3 1 2 3 12I I 2 I 5I 1 I I I 0 ! = " # ! + = \$ # + + = % 1 2 3 I =0.143 I 0.286 I 0.143 mA mA mA = ! = A 2 B 1 V (1V)+(1k )I 2 2.714 V (1V)+(2k )I 1.286 V V = ! + = + = ! = + 2. [12 pts total, 2 pts each part each circuit] Consider the circuit below. a. Determine if the diode is ‘on’. Assume the diode turn-on voltage is 0.7 V. b. Find the currents I 1 , I 2 and I D in each circuit. c. Find the voltage at node X in each circuit. Circuit 1 Circuit 2 I 1 I 1 I 1 I 2 I 3 I 4

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 4 Solutions Capacitors and Resistor Circuits (plus Diodes) Page 2 of 12 Circuit 1: a) We assume the diode is ‘off’ and replace it by an open circuit in the equivalent circuit we use to find the voltage across the diode. Since I D =0, I 1 =I 2 =I KVL: 5 ! 3I ! I+3=0 I= 8V 4k " = 2mA KVL: 5 ! 3I ! V OC ! 2I D = 0 V OC = 5 ! 3I=5 ! 3 # 2 = ! 1V Since –1V < +0.7V, the diode is indeed ‘off’ The assumption was correct b) I 1 = I 2 = 2mA, I D = 0 c) X V 5 (3k )I=5 3 2mA= 1V = ! " ! # ! Circuit 2: a) The diode is flipped, otherwise it is the same circuit. We must remember to define the diode voltage as shown below. V D + ! anode cathode Since the equivalent circuit is the same as above, V D = +1V > 0.7V The diode is ‘on’. The assumption was wrong! Need to re- analyze! b) We note that when the diode is ‘on’, its voltage is 0.7V, D 1 2 I 0 I I ! " ! . KVL 1: 5 ! 3I 1 + 0.7 – 2I D = 0 KVL 2: 3 + 2I D ! 0.7 ! I 2 = 0 KCL: I 1 ! I 2 ! I D = 0 " # \$ % \$ We have 3 equations and 3 unknowns. I 1 =1.973mA, I 2 =2.082mA, I D = –0.109mA c) X 1 V 5 3I 5 (3k )(1.973mA) 0.918V = ! = ! " = !
Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 4 Solutions Capacitors and Resistor Circuits (plus Diodes) Page 3 of 12 3. [16 pts total, 4 pts each] The capacitor in the circuit shown below is uncharged prior to closing the switch at t = 0. a. What is the time constant for capacitor charging in seconds? b. How long does it take for the capacitor to reach 6 V after closing the switch? c. What is the capacitor current when the capacitor voltage is 6 V? d.What is the magnitude of the stored charge at that instant? a) 3 6 3 -4 RC=10 100 10 F =10 10 0.1s ! "# # # = b) -t/0.1 6=10(1-e ) => -t/0.1 0.6=1-e => -t/0.1 e 1 0.6 0.4 = ! = So, t= 0.1ln(0.4)=0.092s ! c) C dv i C dt = can be used when -t/RC C O v V (1-e ) = . An easier method is to find the current flowing through the resistor. Since they are in series, this is also the capacitor current. When v C =6V, v R =4V (to satisfy KVL) i R ( t ) | v C ( t ) = 6 V = 4 V 1 k ! = 4 mA = i C ( t ) d) Q = Cv C = (100 μ F )(6 V ) = 6 ! 10 " 4 Coulombs

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 4 Solutions Capacitors and Resistor Circuits (plus Diodes) Page 4 of 12 4. [15 pts total, 3 pts each] The switch in the circuit below is initially closed . The capacitor begins to charge when the switch is opened at t = 0.
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HW4_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 4...

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