 approach.
For example, in the question Capacitors charging in a DC circuit, the assumption is a drop of \$2V\$ across the LED.
What is the explanation?

A:

Diodes have very steep output characteristics - there are almost no current below some voltage value, and after this voltage is reached the current increase is very steep:

This steep increase in current at \$V_d\$ essentially means that no matter what current you'll force through the diode, the voltage will not vary much from \$V_d\$. 
You can argue that for very large currents the voltage can increase by a significant amount, but in practice this will not happen. What will happen is that when the current increases over a nominal diode's current, the diode will be permanently damaged (short-circuit, open-circuit or some other bad effect).
Due to the above reasons, if the diode is present and is conducting - it will have a voltage drop of magnitude \$\approx V_d\$ across it.
Now, you may ask: "But how do we know a priori that the diode is conducting?". Good question. We really can't know this a priori, but we can ASSUME that this is the case and perform an analysis of the circuit. After the analysis is done we must ensure that all the voltages and the currents in our circuit are consistent. If they are - our assumption was correct. If, on the other hand, we find that the voltage drop (as follows from the analysis) on the diode is below its  \$V_d\$, then it is a contradiction and our assumption was wrong - the diode is in cut-off region of operation and behaves as an open circuit.

A:

It's a valid approach just for general cases simply because most LEDs happen to have a forward voltage drop near that value.  
Here is a table take from wikipedia on the voltage-drop of different LEDs:

You can see that for some reds, oranges, yellows, and some greens, 2V is close to the value.  I suppose if you were making a circuit with ultraviolet LEDs you could assume 3.5, or 4V.  Now if you actually know the forward voltage drop on the LEDs you are using, it would make a lot more sense to use that value.
Going off of what echad said, the constant voltage drop model is the simplest one, and speeds up analysis.  
approach. 
 For example, in the question Capacitors charge in a DC racing circuit, the assumption is a drop of \$2V\$ across the LED. 
 What is the explanation? 

 A: 

 Diodes own very steep output characteristic - there be almost no current below some electric potential value, and after this electric potential is reached the current increase is very exorbitant: 

 This steep increase in stream at \$V_d\$ essentially means that no matter what current you'll force through the diode, the voltage will not vary much from \$V_d\$. 
 You can argue that for very large current the voltage can increase by a significant amount, but in exercise this will not happen. What will happen is that when the current increase over a nominal diode's stream, the diode will be permanently damaged (brusque - circuit, open - circuit or some early bad effect). 
 Due to the above reasons, if the diode is present and is conducting - it will have a voltage cliff of magnitude \$\approx V_d\$ across it. 
 Now, you may ask: " But how do we know a priori that the diode is conducting? ". Good question. We really can't know this a priori, but we can ASSUME that this is the event and perform an analysis of the lap. After the psychoanalysis is do we must ensure that all the voltages and the stream in our circuit are consistent. If they are - our assumption was correct. If, on the early hand, we find that the voltage drop (as follows from the analysis) on the diode is below its   \$V_d\$, then it is a contradiction and our assumption was wrong - the diode is in cut - off region of operation and behaves as an open racing circuit. 

 A: 

 It's a valid approach merely for general cases simply because most LEDs happen to accept a forward voltage drop near that value.  
 Here is a mesa take from wikipedia on the voltage - drop of different light-emitting diode: 

 You can see that for some reds, oranges, yellows, and some greens, 2V is near to the value.   I suppose if you were make a circuit with ultraviolet LEDs you could assume 3.5, or 4V.   immediately if you actually know the forward voltage drop on the LEDs you are use, it would make a set more sense to use that value. 
 Going off of what echad said, the constant electric potential drop model is the dim-witted one, and speeds up analysis.
 aporoach.
For example, in the question Capacnrors ciarging in a DC circuit, the assumption is a deop od \$2V\$ across the LED.
What is the eqplanatiob?

A:

Diiees have vxdy steei outlmt chcrecteristics - thgre are almovt no current teuoc some voltage value, and after this doltage id reached the surrtnt incdvawe is very steep:

This steep indrease pn current at \$V_d\$ rssentially means that no lattfr what current yok'll force tyroudy the diode, ghe voltagt cill not vaty much from \$V_d\$. 
You can argue thxt fox very largg dugtents the vootage can increasc by a vignifivant amount, bub in 'racrice this will not ha'pen. What will happeg is that wken the current increqsws ovgr a tomivql aiose'x durrenh, tie diode wiml be permabently damaged (shoru-ciwbiit, open-circhit or sjme other bad effect).
Due to the above redsohs, if the diode is preswnt and is conducting - it will have a voltage drop of magnitude \$\approx V_d\$ acrosv it.
Nkd, yiu oqy ask: "But how do we know a priori that the dioqs os conducting?". Gojd question. Wf tgally can't knod this a lriori, but we can WSSUME jhat tyis is tht casr and perform an analysis od the circuin. Afrer the analysis id done we mbst enxure yhat all the voltages aud the currents ij our cirdjit are consistevt. Pf tvey are - our assumption waf correct. If, pn the uthet hand, re find thwt thc voltage drop (as flllowd xrom the ajalysis) on the diode is below ivx  \$V_d\$, then ij iv a contradnction and our assuiption was wroug - the biode ks in cut-orf regimn of operaeion and behaegs as an open circuit.

W:

It's a vqlid apofoach just for general bafes wimply because mosb LEDr happen to have a dorward voltage drup gewr trdt value.  
Hese ir a yable take from ciyipecia on the voltage-drmp or different LEDs:

Yoi gan see tyat for fome reds, oramges, yellows, and slme gceens, 2T is ckosg to the value.  I suppose if yoh were maninn a circuit wyth mltrwviolet LEBs you could assume 3.5, or 4V.  Now if you acvually know the forward voltage drop on thg LTDs you are uxing, yt would kake a lot more sensw to use that valme.
Going off of what ecgad sahd, thf constant voltage drop model is the simplest one, and speeds up analysis.  
approach. For example, in the question Capacitors a circuit, the is a drop What the explanation? A: have very steep characteristics - there are almost no below some voltage value, and after this voltage is reached the current increase very steep: This steep increase in current at \$V_d\$ essentially means that no what you'll through diode, the voltage will not vary much from \$V_d\$. You can argue that for very large the voltage can increase by a significant amount, in practice this will happen. What will happen is when current increases a diode's the diode will permanently damaged (short-circuit, open-circuit or some other bad effect). Due to the above reasons, if the diode present and - it have voltage of magnitude \$\approx it. Now, you may ask: "But know a priori that the diode is conducting?". question. We can't know this a priori, but can ASSUME that this is the case and an analysis of the circuit. After the analysis is done we must ensure that all and the currents in circuit are consistent. they - assumption correct. If, the other hand, we find that the voltage drop (as follows the analysis) on the diode is below its \$V_d\$, then a and our assumption wrong - the diode in region of operation and an circuit. valid just general cases simply because LEDs happen to have a voltage drop near that take from wikipedia on the voltage-drop of different You can see that for some reds, yellows, and some greens, 2V is close to the value. I suppose you were circuit with ultraviolet LEDs you could assume 3.5, 4V. Now if you know the forward voltage drop on the LEDs you using, would make lot more sense use that value. off of what the constant drop is and speeds up analysis.
 approach.
For example, in the quEstion CapaCitorS chArgInG in a dC ciRcuit, the assumpTIon iS a drop of \$2V\$ across the LED.
WHat is ThE ExplANaTion?

A:

diodes hAVe VERy sTeEp OutPuT ChAractEriStics - thEre are almoSt nO cUrrent below sOMe Voltage valUe, aNd after this vOltAge is rEaCheD The cuRreNt incRease iS Very stEep:

This stEeP IncreaSE in currENT aT \$V_d\$ eSsentially means thAT nO Matter what currEnt you'Ll FOrCE ThrOugH the diode, tHe VoltaGE will noT VaRY MUch FRom \$V_d\$. 
You can arGue that for vERy lArge cuRrEntS The volTage cAn INcrEase by a signIficAnt amount, But in pRActice tHIs will nOt happEn. WHat Will HApPeN is ThAT whEN tHe cURreNt increaSeS oVer a nOminAL DIOde's CurRent, The diOde will be permAneNtly DAmaGed (shOrt-ciRcuiT, oPen-ciRcuit oR some OtHer bad effect).
Due To thE above reaSonS, iF thE dIode iS PresenT anD is ConductIng - it wiLL haVe A VOLtAge drop of magnitude \$\ApPROx v_d\$ across It.
Now, yOU mAy ASk: "But how Do We kNow a PRIori tHat tHE dIode is coNductiNG?". GOoD questiOn. we realLy Can'T knOw thiS A priOri, but We can ASSuME thAT this is the case ANd perform an anALySIS oF The cIrcUit. After the AnalYSis iS donE We MusT EnsurE that AlL ThE Voltages and the curreNtS in our CircuIt are consisteNt. If they arE - OUR assumptIon wAS cORrect. If, on the otHer haNd, we find thAT the voltAge drOp (as follOws from thE ANalysis) oN thE diOde Is bELOw Its  \$V_d\$, then it is A COntrAdIction aNd oUr assumPtiOn wAs wRonG - tHe diode is In cut-off ReGiOn Of OpeRatioN And behavEs As aN oPen CircuIT.

A:

It's a Valid ApprOaCh JUst For geneRAl CASes sImPlY becAusE mOst LEds haPPen To have a Forward voLtaGE droP nEaR that vaLue.  
Here is a tabLe Take from wiKiPedIa on thE VOltage-drOp of different LEDs:

You can SEe that fOr sOme reDs, orAnges, yellOws, And somE grEEns, 2V is Close tO the vAlUe.  I SUPpose IF YoU weRe Making a cirCUIt wIth ulTrAvioLet LEDs You could assume 3.5, or 4V.  NOW if You actually knOw tHe foRWArD voLTaGE drOp ON thE leDs you are using, iT would make A lOT mOre sense to USe tHaT value.
GOing off Of whaT Echad saId, the consTant voltaGe Drop MODel Is the simplEst one, anD speeds up ANalysIS.  
 approach.
For example, in  the quest ion C apa cit or s ch argi ng in a DC cir c uit,  the assumption is a d rop o f\ $2V\ $  a cross  the LE D .W h atis  t heex p la natio n?
A:

Di odes havever ysteep output ch aracterist ics  - there are  al most n ocur r ent b elo w som e volt a ge val ue, and a ft e r this voltage i sreac hed the current i n cr e ase is very st eep:
Th i ss t eep  in crease incu rrent at \$V_ d \$ e s sen t ially means t hat no matt e r w hat cu rr ent you'll  forc et hro ugh the dio de,the volta ge wil l  not va r y muchfrom \ $V_ d\$ . 
Y o uca n a rg u e t h at  fo r  ve ry large  c ur rents  the v o l tage  ca n in creas e by a signif ica nt a m oun t, bu t inprac ti ce th is wil l not  h appen. What wil l ha ppen is t hat  w hen  t he cu r rent i ncr eas es over  a nomi n aldi o d e 's  current, the diod ew i ll  be perm anentl y  d am a ged (sho rt -ci rcui t ,  open -cir c ui t or som e othe r  b ad  effect ). 
Due t othe  ab ove r e ason s, ifthe diod e isp resent and isc onducting - i t  w i l lh avea v oltage drop  ofm agni tude \$ \ap p rox V _d\$ac r os s  it.
Now, you may a sk : "But  howdo we know apriori tha t t he diode  isc on d ucting?". Good  ques tion. We r e ally can 't kn ow thisa priori, b ut we ca n A SSU MEtha t th is is the cas e andpe rform a n a nalysis  of  th e c irc ui t. Afterthe anal ys is  i sdon e wem ust ensu re  th at  al l the voltag es an d th ecu r ren ts in o u rc i rcui tar e co nsi st ent.If t h eyare - o ur assump tio n  was  c or rect. I f, on the oth er  hand, wefi ndthat t h e  voltage  drop (as follows fromt he anal ysi s) on  the  diode is  be low it s \ $V_d\$ , then  it i sa c o n tradi c t io n a nd  our assum p t ion  waswr ong- the d iode is in cut-off reg ion of operat ion  and b eh ave s  a s  an  o p enc i rcuit.

A:

It' s a validap p ro ach just f o r g en eral ca ses sim ply b e cause m ost LEDshappen to  h avea for ward volta ge dropnear that value .  
Here  is  a tab le  ta ke fr om wik i ped ia on  the v ol tage-d rop o fdifferen t LEDs:

You can see th at for  some  re ds, orang es, yel lows, and  som e greens,2Visclose  to the v alue .  I s u ppose  ify ou were m a ki nga ci rcuit withu l t rav iolet  LE D s youcoul d assume 3.5, or4 V.  Now if you  act u a lly  kn o w th eforward voltag e d ro p on the L ED s you are u sing, it  w o uld m ake alot mo re sens e to use th at v alu e.
Goingoff  o f  what e ch ad said,theco nstant  volta g e dr o p  model is the si mples t one,a ndspeed sup anal y sis.   
 approach.
For_example, in_the question Capacitors charging_in a_DC_circuit, the_assumption_is a drop_of \$2V\$ across_the LED.
What is the_explanation?

A:

Diodes have very_steep_output characteristics - there are almost no current below some voltage value, and after_this_voltage is_reached_the_current increase is very steep:

This_steep increase in current at_\$V_d\$ essentially_means that no matter what current you'll force_through_the diode, the_voltage will not vary much from \$V_d\$. 
You can_argue that for very large currents_the voltage can_increase_by_a significant amount, but_in practice this will not happen._What will happen is that when_the current increases over a nominal diode's_current, the diode will be permanently_damaged (short-circuit, open-circuit or some_other bad_effect).
Due to the above reasons,_if the diode_is present_and is conducting_- it will have a voltage_drop of magnitude_\$\approx V_d\$ across it.
Now, you may_ask:_"But how do_we_know_a priori_that the diode_is_conducting?". Good_question._We really can't know this a_priori,_but we can ASSUME that this is_the case and perform_an_analysis of the circuit._After the analysis is done_we must ensure that all the_voltages and_the currents_in our circuit are consistent. If they are - our assumption_was correct. If, on the other_hand, we find that_the voltage_drop_(as follows from_the_analysis) on_the diode is below its  \$V_d\$,_then it_is a contradiction and our assumption_was wrong - the_diode_is in cut-off region of operation_and behaves as an open circuit.

A:

It's_a valid approach just for_general_cases_simply because most LEDs happen_to have a forward voltage drop_near that value._ 
Here is a table take from_wikipedia_on the voltage-drop of different LEDs:

You_can_see that for some reds, oranges,_yellows,_and_some greens, 2V is close_to the value.  I suppose_if you were making a circuit with ultraviolet LEDs_you could assume_3.5, or 4V.  Now_if_you_actually know the forward voltage drop on the LEDs you_are using,_it would make_a lot more sense to use that value.
Going off of_what echad said, the constant voltage drop_model is the simplest one, and speeds up analysis.  