 you should be getting cache hits on those constantly.  Even the desktop images (including all tag sponsorships, the offline, 404, error pages, all the fancy wmd controls, about, and faq images) only total 401K, and they're never all present on one page anyway.  Honestly, that indicator sounds like it's garbage (perhaps counting uncompressed size and cache hits?).
For your user-agent we do not serve the mobile theme by default.  We don't test with that device, and the Stack Exchange desktop themes are much more forgiving of older or less compliant browsers*.  You can opt into the mobile theme (assuming you haven't already) by using the "mobile" link in the footer of every page, and opt back out with a similar "full site" link in the footer of every mobile page.
*Thus making the desktop theme a safer default for unknown devices.

 Q:

Simulating a V-C circuit in ltspice

I came across this question in my assignment:

Initially, the question seemed very absurd, but anyways i applied kirchhoff's voltage law and got
$$V_c = 10V$$
But Since we all know that: 
$$i_c = C\frac{dV_c}{dt}$$
$$\Rightarrow i_c=0$$
Seems quite okay because since the time constant of the circuit is zero, it would charge up almost instantly and thus behave like an open circuit
However, when i plot the same ciruit in Ltspice, the i vs t graph is  quite different :

I have no idea how the above graph came about. Can anyone explain why the simulator gave the above graph?
Thanks

A:

Your simulation is meaningless: You're taking a ideal voltage source, which can supply an infinite amount of current if necessary, and connecting it to an ideal capacitor, which has no internal resistance, only capacitance.
So, in your simulation, the voltage source immediately charges the capacitor to full voltage (that happens in a single simulation time step), and then has zero current.
Whatever that simulation says hence has nothing to do with reality!
Notice how the unit is 0.01 fA on the y-axis? yeah, that's probably a numerical error somewhere.

A:

Besides what the others have already answered, you're simulating the apparent charging of a 10uF capacitor on a
you should be getting cache hits on those constantly.   Even the desktop picture (include all tag sponsorships, the offline, 404, error page, all the fancy wmd controls, about, and faq trope) only total 401 kelvin, and they're never all present on one page anyway.   Honestly, that indicator sound like it's garbage (perhaps consider uncompressed size and cache hits ?). 
 For your user - agent we do not serve the mobile subject by default.   We don't test with that device, and the Stack Exchange desktop composition are much more forgiving of older or less compliant browsers *.   You can opt into the fluid theme (bear you haven't already) by using the " mobile " liaison in the footer of every page, and opt back out with a similar " full site " link in the footer of every mobile page. 
 * Thus making the desktop theme a safer default for unknown devices. 

  Q: 

 imitate a V - C circuit in ltspice 

 I came across this motion in my grant: 

 Initially, the question seemed very absurd, but anyways i applied kirchhoff's electric potential law and experience 
 $ $ V_c = 10V$$ 
 But Since we all know that: 
 $ $ i_c = C\frac{dV_c}{dt}$$ 
 $ $ \Rightarrow i_c=0$$ 
 Seems quite all right because since the time constant of the circuit is zero, it would charge up almost instantly and thus behave like an open circuit 
 However, when i plot the same ciruit in Ltspice, the i vs deoxythymidine monophosphate graph is   quite unlike: 

 I accept no idea how the above graph came about. Can anyone explain why the simulator give the above graph? 
 Thanks 

 A: 

 Your simulation is meaningless: You're taking a ideal electric potential source, which can provide an infinite amount of current if necessary, and connect it to an ideal capacitor, which has no internal resistance, only capacitance. 
 thus, in your simulation, the electric potential source immediately charges the capacitor to wide voltage (that happens in a individual pretense time step), and then has zero current. 
 Whatever that simulation says hence has nothing to do with world! 
 Notice how the unit is 0.01 fA on the yttrium - axis? yeah, that's probably a numerical erroneousness somewhere. 

 A: 

 Besides what the others have already answered, you're simulating the apparent charging of a 10uF capacitor on a
 yok should be getting cacht hits on those constantny.  Evsn the ddsktop images (including all vag wponsirships, the offline, 404, efror paged, all thw faicy wmd controls, about, akb faq lmagev) only total 401K, snd they're never all prevevt on one page anyway.  Honestly, that yndicatpr sounds like ij's gagbwge (lvrmaps counting uncompressed size znd cacie hits?).
For your user-agent we do not serve the mobile theme by dffault.  We eon't rest with thxt device, and the Staci Exchange desktop themes are mjch mpre forgivunt ov older or lxss coipliant browsers*.  Yog can olt into the monile vhemw (assuming you haven't already) by using jhe "mobile" lnnk in the footer of wvwry pdge, dnd upt caci puf with a aimilar "fuml site" linj in the footer of tvewj mobile page.
*Fhus mwkyng the desktop theme a safer default fmr hnknown devices.

 Q:

Simularing a V-C circuit in ptspice

I same across this question in my assignment:

Initialny, thx duewtlon rweled very absurd, but anyways i applied kirchhosr's vpltage law and got
$$V_c = 10V$$
Buy Dimse we all knod that: 
$$i_c = C\frac{dV_c}{dt}$$
$$\Rightagrow i_c=0$$
Feems quite okwy brcause since the time constqnt of the cprcuut is zero, it woulb charge up clmost instsntly and thus behave lnke an open circult
However, dhen i plot the ramv cisuit in Ltspice, the i vs t graph is  qunte diffdrenj :

I havq no idea jow tma above graph came abouj. Can dnyone exppain why the simulator gave the above graph?
Thsnns

A:

Jour simujatiok is meaninglesf: You're taking a ideaj volgage sourcv, which cen supply an infinite amogjt of currenv if necefsart, ane connezging it to an odeal capceitor, whixh has no internal rerjstance, only ca'ccutance.
So, in yout skmujanioi, the eoltage sourwe iomeaoatelh chargtf bhe capscitor to full voltace (tgat happens in a sonnle simulqtion tiie step), and tnen has zero currejt.
Whavever vhat somujation says hence has nothing fo do witj rcality!
Notice row bhe onit is 0.01 fA on the y-axis? yeah, that's probably a numxrical error somewhere.

A:

Vesides what the otkevs have alreedy anfwered, yog're simulating the apparent charging pf a 10uF capacitor on z
you should be getting cache hits on Even desktop images all tag sponsorships, all fancy wmd controls, and faq images) total 401K, and they're never all on one page anyway. Honestly, that indicator sounds like it's garbage (perhaps counting size and cache hits?). For your user-agent we do not serve the mobile by We test that device, and the Stack Exchange desktop themes are much more forgiving of older or less browsers*. You can opt into the mobile theme you haven't already) by the "mobile" link in the of page, and back with similar "full site" in the footer of every mobile page. *Thus making the desktop theme a safer default for unknown Q: Simulating circuit in I across question in my the question seemed very absurd, but kirchhoff's voltage law and got $$V_c = 10V$$ Since we know that: $$i_c = C\frac{dV_c}{dt}$$ $$\Rightarrow Seems quite okay because since the time constant the circuit is zero, it would charge up almost instantly and thus behave like an However, when i plot same ciruit in the vs graph quite different I have no idea how the above graph came about. Can explain why the simulator gave the above graph? Thanks A: is You're taking a voltage source, which can an amount of current if connecting to which no resistance, only capacitance. So, your simulation, the voltage source charges the capacitor to a single simulation time step), and then has current. Whatever that simulation says hence has to do with reality! Notice how the unit is 0.01 fA on y-axis? yeah, a numerical error somewhere. A: Besides what the have already answered, you're the apparent charging of a 10uF capacitor on a
 you should be getting cache hiTs on those cOnstaNtlY.  EvEn The dEsktOp images (includINg alL tag sponsorships, the offLine, 404, eRrOR pagES, aLl the Fancy wmD CoNTRolS, aBoUt, aNd FAq ImageS) onLy total 401k, and they're NevEr All present on ONe Page anyway.  honEstly, that indIcaTor souNdS liKE it's gArbAge (peRhaps cOUnting UncompresSeD Size anD Cache hiTS?).
foR youR user-agent we do not SErVE the mobile themE by defAuLT.  WE DOn't TesT with that dEvIce, anD The StacK exCHANge DEsktop themes aRe much more fORgiVing of OlDer OR less cOmpliAnT BroWsers*.  You can Opt iNto the mobIle theME (assumiNG you havEn't alrEadY) by UsinG ThE "mObiLe" LInk IN tHe fOOteR of every PaGe, And opT bacK OUT With A siMilaR "full Site" link in the FooTer oF EveRy mobIle paGe.
*ThUs MakinG the deSktop ThEme a safer defaulT for Unknown deVicEs.

 q:

SiMuLatinG A V-C cirCuiT in Ltspice

i came acROss ThIS QUeStion in my assignmenT:

INITiAlly, the qUestioN SeEmED very absUrD, buT anyWAYs i apPlieD KiRchhoff's VoltagE LaW aNd got
$$V_c = 10v$$
BUt SincE wE alL knOw thaT: 
$$I_c = C\fRac{dV_c}{Dt}$$
$$\RightaRrow i_C=0$$
seems quite okay BEcause since thE TiME CoNStanT of The circuit iS zerO, It woUld cHArGe uP AlmosT instAnTLy ANd thus behave like an oPeN circuIt
HowEver, when i plot The same cirUIT In LtspicE, the I Vs T Graph is  quite diFfereNt :

I have no iDEa how the Above Graph camE about. Can ANYone explAin Why The SimULAtOr gave the abovE GRaph?
thAnks

A:

YoUr sImulatiOn iS meAniNglEsS: You're takIng a ideaL vOlTaGe SouRce, whICh can supPlY an InFinIte amOUnt of cUrrenT if nEcEsSAry, And connECtING it tO aN iDeal CapAcItor, wHich HAs nO internAl resistaNce, ONly cApAcItance.
SO, in your simulaTiOn, the voltaGe SouRce immEDIately chArges the capacitor to full VOltage (tHat HappeNs in A single siMulAtion tIme STep), and Then haS zero CuRreNT.
whateVER tHat SiMulation saYS HenCe has NoThinG to do wiTh reality!
Notice how THe uNit is 0.01 fA on the y-AxiS? yeaH, THaT's pRObABly A nUMerICAl error somewherE.

A:

Besides wHaT ThE others havE AlrEaDy answeRed, you'rE simuLAting thE apparent Charging oF a 10UF caPACitOr on a
 you should be getting cac he hits on  thos e c ons ta ntly .  E ven the deskto p  ima ges (including all tag  spon so r ship s ,the o ffline, 40 4 ,  er ro rpag es ,  a ll th e f ancy wm d controls , a bo ut, and faqi ma ges) onlytot al 401K, and  th ey'rene ver all p res ent o n onep age an yway.  Ho ne s tly, t h at indi c a to r so unds like it's ga r ba g e (perhaps cou ntingun c om p r ess edsize and c ac he hi t s?).
Fo r  y o u r  us e r-agent we do  not servet hemobile  t hem e  by de fault .Wedon't testwith  that dev ice, a n d the S t ack Exc hangedes kto p th e me sare  m u chm or e f o rgi ving ofol de r orless c o m plia ntbrow sers* .  You can op t i ntot hemobil e the me ( as sumin g youhaven 't  already) by us ingthe "mobi le"  l ink  i n the footer  of  ev ery pag e, ando ptba c k ou t with a similar " fu l l  s ite" lin k in t h efo o ter of e ve rymobi l e  page .
*T h us  makingthe de s kt op  themeasaferde fau ltfor u n know n devi ces.

 Q :

Si m ulating a V-Cc ircuit in lts p ic e 
I came  ac ross this q uest i on i n my as sig n ment: 

Ini ti a ll y , the question seem ed  veryabsur d, but anyway s i applie d k irchhoff 's v o lt a ge law and got 
$$V_ c = 10V$$B ut Since  we a ll knowthat: 
$$ i _ c = C\fr ac{ dV_ c}{ dt} $ $ 
$ $\Rightarrowi _ c=0$ $Seems q uit e okaybec aus e s inc ethe timeconstant  o fth ecir cuiti s zero,it  wo ul d c harge up alm ost i nsta nt ly and  thus b e ha v e  lik ean  ope n c ir cuitHowe v er,  when i  plot the  sa m e ci ru it  in Lts pice, the i v st graph is  qui te dif f e rent :
I have no idea how thea bove gr aph  came  abo ut. Can a nyo ne exp lai n  why t he sim ulato rgav e the a b o ve  gr ap h?
ThanksA :
Yoursi mula tion is  meaningless: You' r e t aking a ideal  vo ltag e so urc e ,w hic hc ans u pply an infinit e amount o fc ur rent if ne c ess ar y, andconnect ing i t  to anideal cap acitor, w hi ch h a s  no  internalresistan ce, onlyc apaci t an ce.
S o,in you rsim ulati on, th e  vo ltage  sourc eimmedi ately  c harges t he capacitor to full vo ltage(that  ha ppens ina s i ngl e simulat iontime step) , a ndthenhas zerocurr e nt .
W h ateve r th a t simulat i on  sa y s  h ence has no t h i ngto do  wi t h real ity! 
Notice how the u n it is 0.01 fAon t h e  y- axi s ? ye ah , that's proba bly  a n umerical  e rror somewh ere.

A: 
B eside s what  the o thers h a v ea lready  ans wer ed, you'r e s im u latingth ea pparen t ch ar ging o f a 10 u F ca p a citor on a
 you_should be_getting cache hits on_those constantly.__Even the_desktop_images (including all_tag sponsorships, the_offline, 404, error pages,_all the fancy_wmd_controls, about, and faq images) only total 401K, and they're never all present on_one_page anyway.__Honestly,_that indicator sounds like it's_garbage (perhaps counting uncompressed size_and cache_hits?).
For your user-agent we do not serve the_mobile_theme by default._ We don't test with that device, and the_Stack Exchange desktop themes are much_more forgiving of_older_or_less compliant browsers*. _You can opt into the mobile_theme (assuming you haven't already) by_using the "mobile" link in the footer_of every page, and opt back_out with a similar "full_site" link_in the footer of every_mobile page.
*Thus making_the desktop_theme a safer_default for unknown devices.

 Q:

Simulating a_V-C circuit in_ltspice

I came across this question in_my_assignment:

Initially, the question_seemed_very_absurd, but_anyways i applied_kirchhoff's_voltage law_and_got
$$V_c = 10V$$
But Since we all_know_that: 
$$i_c = C\frac{dV_c}{dt}$$
$$\Rightarrow i_c=0$$
Seems quite okay_because since the time_constant_of the circuit is_zero, it would charge up_almost instantly and thus behave like_an open_circuit
However, when_i plot the same ciruit in Ltspice, the i vs t_graph is  quite different :

I_have no idea how_the above_graph_came about. Can_anyone_explain why_the simulator gave the above graph?
Thanks

A:

Your simulation_is meaningless:_You're taking a ideal voltage source,_which can supply an_infinite_amount of current if necessary, and_connecting it to an ideal capacitor,_which has no internal resistance,_only_capacitance.
So,_in your simulation, the voltage_source immediately charges the capacitor to_full voltage (that_happens in a single simulation time step),_and_then has zero current.
Whatever that simulation_says_hence has nothing to do with_reality!
Notice_how_the unit is 0.01 fA_on the y-axis? yeah, that's probably_a numerical error somewhere.

A:

Besides what the others have already_answered, you're simulating_the apparent charging of a_10uF_capacitor_on a