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**Unformatted text preview: **EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS !"#$%& 1. Hambley, P2.78 % Start by finding Rt , Isc , Voc !"#$%&'()*+",+-"#%+&)."#"&)/0% 1&23")45%6(%7").%3 :; " !9%8 <% !"#$%#$(%/&-,+(%=(,&(.0% Solving for Rt by zeroing out the current source gives: B# " We can see that the 5" ?<>9%%and the 10 resistors are in series. The series combination (on top) % #9 & parallel with the remaining 5 resistor. This gives: @ 9 & @are in@A $ of those two
C$(%(D-"3:2()#%+",+-"#/%:,(0% + 10) = R = 5||(5
t 1 1/5+1/(5+10) @ % = 3.75 Solving for Isc by first shorting out terminals a, b This removes the 5 resistor on the right from the circuit. To calculate Isc now just requires a current divider, with only the upper 5 in parallel with the 10 Isc = -2A 10 10+15 = -1.33A !"#$'& Solving for Voc by opening the terminals a, b E&#"+(%#$(%/&-,+(%'&2:,"#F%,(2:#"3(%#&%#(,G"):2/%:%:).%;<% Now the two 5 resistors are in series, combining them gives us a 10 resistor. This 10 & is in parallel with the original 10. Thus, the 2 A current is split equally between the two. Therefore we can see that 1A of current flows through the 5 resistors. Through Ohm's C$(%@A*%%,(/"/#&,%$:/%)&%(77(+#%&)%#$(%(D-"3:2()#%+",+-"#/%;(+:-/(%#$(% Law we can tell that the voltage drop across a, b is -5V, due to the direction of current flow. 3&2#:4(%:+,&//%#$(%@H*8%/&-,+(%"/%").('().()#%&7%#$(%,(/"/#&,%3:2-(<% Voc = -1A 5 = -5V Note: It is perfectly acceptable to use Voc = Isc Rt , this just demonstrates how to solve for all terms directly % & & !"#%(& &
1 C$(%C$I3()")%3&2#:4(%"/%(D-:2%#&%#$(%&'()*+",+-"#%3&2#:4(%6$"+$%"/%@H<J%8<%% C$(%(D-"3:2()#%+",+-"#%6"#$%#$(%A<@*%%2&:.%+&))(+#(.%"/0% 2 @ EE 40, FALL 2008 PROF. CHANG-HASNAIN B# " " ?<>9%% % @ 9 & @ #9 & @A $ C$(%(D-"3:2()#%+",+-"#/%:,(0% %
HOMEWORK 3 SOLUTIONS Having found all the parameters, we arrive at the following equivalent circuits: &)'*+&,$-'*',$./+ + % E&#"+(%#$(%/&-,+(%'&2:,"#F%,(2:#"3(%#&%#(,G"):2/%:%:).%;<% +++++++++ 2. Hambley, P2.83 & Start by finding Rt , Isc , Voc !"#$'& C$(%@A*%%,(/"/#&,%$:/%)&%(77(+#%&)%#$(%(D-"3:2()#%+",+-"#/%;(+:-/(%#$(% 250'3 " 2 ++++7,;4'$<=+ Because of the dependent sources, we can no longer solve for Rt by just zeroing out the 3&2#:4(%:+,&//%#$(%@H*8%/&-,+(%"/%").('().()#%&7%#$(%,(/"/#&,%3:2-(<% " 62+: 5+ independent sources. ,& 4 ,& (&)'*+&,$-'*',$./+ + " 12 0 " @+? Solving for Isc by shorting out terminals a, b " '3 ! 250'3 " 1+? +
&
Doing so removes the 10 from the circuit. This is clear because the voltage across it is C$(%C$I3()")%3&2#:4(%"/%(D-:2%#&%#$(%&'()*+",+-"#%3&2#:4(%6$"+$%"/%@H<J%8<%% now zero. The entire 20V drop is across the 5 resistor. We can calculate the current, C$(%(D-"3:2()#%+",+-"#%6"#$%#$(%A<@*%%2&:.%+&))(+#(.%"/0% which is also ix through it: ix = Applying KCL at node a gives:
20-0 5 +>B4#+ & .& " 0+$ 5++A>).+ & 4B;#$*.+B(#/+ !"#%(& = 4A + is c = ix - 0.5 ix = 0.5 ix = 0.5 4A = 2A % Solving for Voc by opening terminals a, b F;#G+E15H0=+8#+9'$-+*>#+A>I4#$'$+#D)'4B;#$*/+ '*+&,$-'*',$./+ % % + & +++++++++ % + JA 50'3 " 2 ++++7,;4'$<=+ " 62+: 5+ 3'G)G+",8#(+'.+,F*B'$#-+9,(+B+;,B-+(#.'.*B$&#+#D)B;+*,+*>#+ +(#.'.*B$+ )'*+&,$-'*',$./+ + % 11 12 04" @+& " K5KKK+J + " * ? B3 '3 ! 2C* '3 " 1+? + 50 B4#+ F;#G+E15HL=+8#+9'$-+*>#+A>I4#$'$+#D)'4B;#$*/+ " 0+$ 5++A>).+ #$*.+B(#/+ + ! '3 # 250'3 " 2 ++++7,;4'$<=+ 62 8#+9'$-+4 ,& " 62+: 5+
PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS + EE 40, FALL 2008 7>,(*%&'(&)'*+&,$-'*',$./+ To analyze this circuit apply KCL at node a and Ohm's Law across the 5: + + + '3 " 12 0 " @+? i = 20-voc 3 4 ,& + + 12 ! 4 ,& '3 " +++++++++ 0 + - ix + 0.5 ix = 0 Solving gives voc = 10V A>#$=+8#+>B4#+ From this and isc we can find Rt by: C* " 4 ,& '.& " 0+$ 5++A>).+ Rt = voc /isc = 5 '.& " '3 ! 250'3 " 1+? + voc 10 x 5 *>#+#D)'4B;#$*.+B(#/+ Solving for all the parameters gives the equivalents: + & 3. Hambley, P2.96 !"#$'+ ?.+'.+E(,F;#G+E15H0=+8#+9'$-+*>#+A>I4#$'$+#D)'4B;#$*/+ (a) With only the 2A source activated, the 1A source has been opened, preventing current +
flow through that branch. Thus, the entire 2A flows through the Element A. i2 = 2A so v2 = 2(i2 )3 = 2 23 = 16V (b) With only the 1A source activated, the 2A source has been opened, preventing current flow through that branch. Thus, the entire 1A flows through the Element A. + + i1 = 1A so v1 = 2(i1 )3 = 2 13 = 2V + A>#$+GB3'G)G+",8#(+'.+,F*B'$#-+9,(+B+;,B-+(#.'.*B$&#+#D)B;+*,+*>#+ (c) With both sources activated, from KCL we can see that 3A flows through Element A. A>I4#$'$+(#.'.*B$+ i = 3A so v = 2(i)2 = 2 33 = 54V + Superposition does not apply because Element A has nonlinear relationship between v %4 1&1 " K5KKK+J + + andEi. " * GB3
4. Hambley, P2.100 C* & !"#$(+ ?.+'$+E(,F;#G+E15HL=+8#+9'$-+*>#+A>I4#$'$+#D)'4B;#$*/+ using superposition (a) Replace the dependent source with a 12 V independent voltage source and solve for i1 , First, zero out the 12 V source and find i1,a , the contribution of the current source to i1 . This is simply a current divider, giving 8+4 i1,a = 1 8+4+6 = 2/3 A + Next, zero out the current source and find i1,b . Then the 3 resistors are all in series, M1 giving us the following ohm's law. 12-0 i1,b = - 6+8+4 = -2/3 A Recall that current for a resistor flows from high potential to lower potential. Because 4 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS of the notation chose, the current i1,b is negative. Summing the two currents gives us i1 . i1 = i1,a + i1,b = 2/3 + -2/3 = 0 A (b) Replace the dependent source with a 0 V independent source. Solving for i1 is actually identical to solving for i1,a from part a. Is that clear? This is the same as just zeroing out the 12 V independent source. Thus i1 = 2/3 A (c) With the dependent source as shown and solving for i1 . There are 4 nodes. Start by referencing the bottom node as ground. Next, The positive terminal of the dependent source is known as 2 vs and the node in the middle of the voltage divider is given the name vs . Thus, only the top terminal of the circuit is unnamed. We will name that as v1 . Setting up a KCL equation at v1 gives: 1 - i1 - Using Ohm's law for i1 gives: i1 = Finally, from voltage divider rule: vx = v1 Plugging in and solving gives i1 = 0.4 v1 = 7.2 vx = 2.4 (d) When using superposition, we cannot zero the dependent source in the circuit. We can only zero out the independent source in the circuit. 5. Hambley, P3.11 Using the following: V = Q/C P =I*V W = 1/2 * C*V2 We arrive at: Because of the notation chosen, the current is negative t 1 t v(t) = C 0 i(t)dt + v(0) = 2 106 0 i(t)dt p(t) = v(t)i(t) 1 w(t) = 2 C v 2 (t) = 0.25 10-6 v 2 (t)
4 4+8 v1 -2vx 6 v1 -vx 8 =0 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 5 ! # $ !! "#! $ " !&" (
& & '& ) # $ !! "#! (
& ! !! "$ !! " ( ' %" * !! " ( * &.*- & '& ') & " * !! " ( 0(!50(4,$!15($4(16740#(+8$78(!7(! % & 9(!50($:$!$26("76!2;0($4(<087(2:#( (a) From terminals x, y we can see that on the left the 10F is in series with 15F. 0( On the right the 1F is in parallel to the 5F, and together they are in series with the 6. Hambley, P3.25 !! " % % ' ! #
& 12F. ! Both networks are in parallel with one another and with the 3F. This gives us the $ !! "#! $ " !& " % '& ?* & '& '= >#! $ & % *&&! % following: #
& Ceq = 3 + 1 1/10+1/15 + 1 1/12+1/(5+1) = 13F ?*& & '& '= > % A(@ % (b) From terminals x, y the 10F is in parallel to 8F. Also, the 4F is in parallel to the 5F. Both networks are in series with each other. This gives us: Ceq =
1 1/(10+8)+1/(4+5) + % "$ % &.A! %
'= = 6F + ?*& & '& > % D(BC (
' * Recall for this problem that Capacitors in parallel add, while capacitors in series have the reciprocals add: 1/C = 1/C1 + 1/C2. , !! " % %" * !! " % &.*! * (E( , ?*& & '& '= > % D& ((E( &#%& !$ "$! " 6 EE 40, FALL " *)*#". ,%-" !#***$ "!7. Hambley, P3.47 6 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS Current flowing through a 2-H inductance is given by: iL (t) = 5 e-20t 7 &"*$ ! Using vL (t) = L dI gives: dt vL (t) = -200e20t V !& #$$' ##$% # #$% " ! *#!% ")& !% " "!-' .$ + !#,$$$' ###$%% !"!-' # #$% " ! # #$$' ()& !% " ! (% # #$% #() !%' #$$ " !" & & ! ()(% " % &! & #$$' # #$% !" # (% ! Using pL (t) = iL (t) vL (t) gives: * & !% ")& !# #$% %" pL (t) = (-5e-20t ) (-200e-20t ) = -1000e-40t W ##$$' # #$% !" # #$% !# #$$' "!-' " ! *#!% ")& !%'"# .$% !+ & ,$$$ & 89 #,$$$' # .$% !+ , &%)& !% "&! # ! # #- # .$% !0 , ' &%)& !% "&! # ! # , # .$% # &0 &% #-')& !% "!! ! # #-' # .$% !0 Using w(t) = 0.5 L i2 gives: wL (t) = 0.5 L [iL (t)]2 = 25e-40t J 2!
8. Hambley, P3.63 (a) The two 2 H and the 0.5 H inductors are removed from the circuit by the short circuit from the 1 H to ground. Thus: Leq = 1H 2! !
% &' $
% , * & !% "(% ( )& !$ "
! ,, % ** !%%"(% ( )& !$ " ! "(% & %'' && #$$ , ! % % * ! "(% ( ) !$ " ,$ & ' * & !% "(% & ! '& % EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 7 (b) On the left: the 30 H and 15 H are in parallel, and together also in series with the 2 H. On the right, the 6 H and 3 H are in parallel and together are in series with the 2 H. Both of these two networks are in parallel with each other. Finally, they are all in series with the 1 H. Thus: 1 Leq = 1 + =4H 1 1 +
1 +2 1/30+1/15 1 +2 1/6+1/3 Recall that inductors add like resistors 9. Hambley, P4.9 Before t=0, or t=0- , we have v(t)=0 because the switch is closed, shorting all elements to ground. After t=0, we can write the following KCL equation at the top node:
v(t) R +C dv(t) dt = 1mA Substituting values and simplifying we get: 0.01 dv(t) + v(t) = 10 dt From earlier, we know that the solution for this is of form: v(t) = K1 + K2 exp(-t/RC) = K1 + K2 exp(-100t) We can figure out initial and final conditions for this problem given what we know about capacitors. At t=0+ we know that the voltage across the capacitor cannot change instantaneously, so we have v(0+ ) = v(0- ) = 0 Also, at steady state we know that a capacitor is open, so all the current will flow through the resistor. Giving: v() = 1mA * 10 k = 10 V. From the initial conditions we can solve for K1 and K2 . v() = 10 = K1 v(0) = 0 = K1 + K2 so K2 = -10. Thus we get the following v(t) equation: v(t) = 10 (1 - exp(-100t)), for t > 0. The graph is shown: S1/!.(6&%-/!%74(00!&1/!7%3%7'&%)7/!7%))(&!71%)-/!')0&%)&%)/($06D?!0(!@/! 1%./! . !L % " $ . !L # " $ L ! ! . !L % " $ L $ N K % N A
8 ! S1$0?! N A $ #N K $ #KL ?!%)8!&1/!0(6$&'()!'0! !& $ FALL 2008 PROF. CHANG-HASNAIN .EE "40, KL # KL /23!# KLL& "!5(4!& & L ! HOMEWORK 3 SOLUTIONS ! !
10. Hambley, P4.17 (a) The voltages across capacitors cannot change instantly. Thus, v1 (0+ ) = 100 V and ! v2 (0+ ) = 0 V KKK + ) by just applying KVL and Ohm's Law: We can solve for i(0 i(0+ ) = (b) Using KVL: -v1 (t) + R i(t) + v2 (t) = 0 Using v = 1/C i(t)dt: t t 0 i(t)dt - 100 + R i(t) + 1/C2 0 i(t)dt + 0 = 0 Taking the derivative gives: di(t) 1 1 C1 i(t) + R dt + C2 i(t) =
di(t) dt 100-0 100k = 1mA 1 C1 + 1 R 1 ( C1 + 1 C2 ) i(t) = 0
C1 C2 C1 +C2 (c) The time constant is just the coefficient of the i(t), = R (d) Solving for i(t): i(t) = K1 + K2 exp(-t/ ) Using properties of capacitors and evaluating gives: i(0+ ) = 1 mA i() = 0 Plugging them in gives K1 = 0, and K2 = 1mA Gives the final equation: i(t) = exp(-20t) mA (e) Solving for the final value of v2 (t)
1 v2 () = C2 0 (t)dt + v2 (0+) = 106 0 10-3 exp(-t/0.05)dt + 0 = 103 (-0.05)exp(-t/0.05)| 0 = 50V =50ms ...

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