hw5sol - EE100\/42 Spring 2011 Prof Poolla HW5 Solution P3.13 cfw Therefore cfw cfw cfw 0 cfw cfw cfw Jcfw cfw cfw $ $ cfw The units are A V W and J for

hw5sol - EE100/42 Spring 2011 Prof Poolla HW5 Solution...

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Unformatted text preview: EE100/42 – Spring 2011 – Prof. Poolla HW5 Solution P3.13 {{ Therefore: {{ # " - { { ' 0 " {{ {{ {{ J{{ {{ {{ # $ $ {{ The units are A, V, W and J for current, voltage, power and energy respectively. P3.22 {{ $""" . {{ J{ { 0 {. 0 {{{{ At t=0, J{ { At t=0.5 ms, J{ { 0. $""" . {0{ $""" $""" $""" { { so it is neither absorbing nor delivering the power. J{ { . 0 the capacitor is absorbing the power. #{ 0{ #{ % and because it is positive, P4.11 (a) After t=0 we have VC=VR and from KCL: - 0 {{ $ 0 #"" % And we know that Vc(t=0)=10 V. Therefore A=10. {{ #"" {{ $"" JJ {{ JJ 2 $"" {{ JJ (b) #"" #"" {{ J{{ J{{ JJ 2 JJ (c) ∞ " ∞ J{{ " $"" . # $"" # $"" $"" ∞ É" H (d) The initial energy store in the capacitor is: # $ ${ # $ { 0 0 $ H P.4.22 and therefore i=0) and At steady state the capacitance has no current (because because the switch is open, all the 10mA current has to pass through 1kΩ resistance. So the steady-state value of vc is vc=10mA×1kΩ=10 V. After opening of the switch at t=0: - $ $ {{ % 0 0 0 0 % $ . #"" JJ 2 99% of the steady-state value of vc is 99%×10=9.9 V. Therefore: . . #"" # Ž #"" %% J P4.27 The circuit has been connected for a long time. So it is in the steady-state condition and we can replace the capacitance with open-circuit and the inductance with short-circuit. Because the capacitance has zero current, all the 4mA current of the current source has to pass through the resistance and inductance and since the inductance in steady-state is short-circuit, iR=4 mA. Writing KVL around the outer loop (resistance, inductance, voltage source and capacitance): 0 -- . P4.28 Before t=0, the circuit is in steady-state and switch is closed. The capacitance current is zero so all the 5mA current should divide between two 10kΩ resistances. Each resistance will have 2.5mA and therefore vC(before t=0)=2.5mA×10kΩ=25V. A long time after t=0, the circuit again is in steady-state condition however this time, the 10kΩ resistance in series with the switch is not in the circuit anymore and its current is zero. So vC(long time after t=0)=5mA×10kΩ=50V . The time-constant after the switch is open is: RC=10 ×103×10×10-6=0.1s - #"0#" 0 # # {{ % { -{ . #" JJ 2 IJ {∞{ ...
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