a — Became fe of vaistanas ARS. ate Connected in suwer A po of V valle arn a a thown vn ait Tg yun ee emaina ame Hey she hy Ty 2p tmey ov may net chee hehe bs Yemain sam Woh:- T4 there aren’ no of equal Taito, keene + Resistance in Pavabll » oH» Conteder hve Tatoos dfeautanca By ks: 1 a conne ced in Paral ("> Bs Apo of V vedi te = Griied as skrun indy Ze ay fe be P-D VW’ Temaing same + Timay or may nor 4 Yemain Same 7 Erte 2% ee deed AL ea a ee Ee tee Be RRR nams- Tanrauranas, Re = BR fa WW is the FeaiAEVINY of th. Makyial : o Ste ae constant] : van aa : : “ah J a for omatal then wit increase It id defined as the Taistana — Asim oltered by a Conductor of unit . nit area 6} Cfotesecton engi, Sr gant area 9} — a Srunit Q-+. Dein scent conte of» condctr and gies a. On win tes doe peaneae ee Kad aa telyee pocal of fhe Yaistane of fered uw be Conductr 9} init Length and Unit ares of crosssecbon 4% ee ee eaten terSiamen mm? z tins. Stneratiee number densify nalite of anata estore cach of resistance R ew will you combine thes to gel f) snaxinen ‘ol In seria tn paral Bmax = 8 Rain = & * & Renae = DE Renae Rimi = NPE Bmin Bnjl Rp s Bude = an ia” 78 7iNe@ Tot Fe + b as sas Tet vaistana swage NA 4 Rap Hae > Pape ib am Bm (6) We smbnion conned xe oe 12 and mei intra aes ‘hen he patent rap arse each eso: tah a) Ree Rie tases 3 eves ming ci ee Sea Lev} ale dea not changeTeststance bf C and Dis the same ad belween Rand 8, Then fhe CqMivalent Circuit Bs kat = 8 ¢ ti ee is Ross B+ ie! >) 12,050 Rape Bird 2 es but @ap = R D He © he Bove i Ba fp a. “Rea(en) e ee wi A(R = he 2e02 ia Rake Rezere R?-28-2-0 Re bs (oraae F zo Ra a4 feogee Eat 2 Re tsa Re ee 1-VEax the capacitor as shown in the circu Find the charge [CBSE (F) 2014] ad 8 ful ed capautees offer 3 inte pond ane for de ny and ze Pn 1or20 Soa Me Fyn Sie C4] Veapacily = 2V Vag = Ex kag canoe 8 Axo Qs cv: Brio bx 2 5 ry o Naps lo say Qs uate 3 f ire 9 4 de ‘ at heal Ioncrssiatanes Alietched ty thrice ia Jengin-whs Ned Yess vity and (1) mew resistance aw wo ee ee a Ws Sxat % but Re lon vel 2s Amd sand aad satel T° Sr ® Wire haw a re ints a wire 0 The Meld Wire fol kts sala tance ofa let is meld and drain half it te nghh: Calcabal the vaissana of lahat it the Wpeccontage change in tex dtana i + “gt mm tas stan, 2 as ia = thet ane sald Bea 2h cyuindricad ive 14 AMretched fe increase I 10 ca fhe Pesce ep oe a ecient ah As Latoit oie de wt Regn! f-rneteade in restetane a ate Daa“ferneteate in festetane = BE x10 e N= 8 x 100 & = 1B 8 x100 ar a gai xiao :21/ wae “Meas dulve complicated A a + ohms Law 8 [ simpleeiewl —> onelbadlery / one current. (Sun cbion ide) mm ae Tae A OAR entering Tundion’ is equal f Tne alg t brace sum of the cur rend Neaving fhe Suackon tas Ty ty Tt obeys Law of Conservation of Charge : (oop rule} 2 we are 4oing In the direchion of Eurrént then T= +Ve f «Hare geing opposite B iancadonol cetrent + TP ye Teach sve ermal oh baller] Tan Veove + TH we Teoun eve fern of Wn balhery then, We =e ehameh +0-4%T +04xE -2 20 : oan ee ‘Apply KUL So Joop BC DAB s e ip 2 meaa Ferly nv. 1s trop c Bape Apply KvL Lo Loop DARED ~baxt +2 -ounr 20 Sotkz 2 40.492 20 ceizeele OL 40-92 22 ye see teh Tah In th ircuit diagram shown here what should be the value of R so that there is no ent in the branch containing 6V battery ? [Ans. 120] -3m 412 125g 46,60 6a lat, 46-0 6, nampa 20 Wm 4 zwar -1-O Aprly RL ts dee DE FED Condit on «From 0-212 Bie Te Pyesary nro Bto-tzs0 > B+6In the network given below, use Kirchhoff's laws to calculate the values of electric ceurrents /,f2 and Aad anely Rew at 0 EAaers Any kui avcen I, o2ftesy) 6 aa8 ai sats 643%, 20 fh 4aas-6-@ amy RU b OC FED atiitjnsst-0 J 4d +2t 12 55320 Byrd aye 13 -O sntnng @ 4 ® : Sn iees ena PR 23ap la MSs ayeue Fam® se 2nut 4 ay shas sik ea a. saat Indy oasige00 ns 6o Shs ig Ti5 sou 11) 6 M97 ede mers 6 ts 66 Parents 3r Apply Kee at F Ips Tyr ta -O Apel UL te loop AFC BA ab hTs +bh ° 6r bt +2=0 ff 32) -2t, +10 A 23-2 Avety ie to roe ‘ee solving ® 4@ neta] +h, - aa Van, ¢12ta+hTa = 8 Iahya tyes = & = 85g 2-3 3x, e4tz -2-@ from® 23) -amL 3g 2Ty4Ta Tye Ore PTasYsa Tense 2019 582.) © Wg s Gime Wh Vers Geta)ne - agp oreo, pyrene Mil Mitaraamceso | “Wnsats Ey az, 1 Ab yega 7 tyra) : Boe] xh A Suth PAG Loy Ws ey Snpwih reriatance £3 the Toate ragant eight ATER cl eo he se ey epance oes ° oe sat evu'be tab bee nn ee chearayerese ry + = ake[ rely eve ‘be love 9 Des -1a42T 4b =o arevi-g From amavisg a fe earev'es 2V eV Ag Find an ex pression Jor (1 potentzal (19 charge 0 RN, The potential drop across the capacitor with steady current in the given ereult will be t+: -O a4 apply evi te tmpavesn JY a Se, Se eae Wa Vietg RAW =O ke Current ta viz 1 e-@ 1-0-0 pf ftv Apply kui fo AmppercD - -2Vie Tear eV'eVe 0 Dye Va (T+ ta] 2esTak= i . (rym jae og From © ©. =Vs [0+ )2e eta k=O the figure. Be [CBSE Allahabad 2015} 4 fel Bee kVL AEE BA Apely Ew ede eae bts 324, 3-6-0 £2-T)R) 210 S(7+t2)+ 4,8) - 10 -O From © zReyee S222 40 perly KVL Bs Abed” Tr 328 +644 - tre = Vad lo axt 2 Vap z + Tt is uked Lo ind the resistance of fre “Yap = av +f any crauit het way twat fone of the or + Circuit Which is in The arm of a Quadritaheal is called os Wheat ston Circuit + MGalvamomeder adi take a bridge b/id the pain Bs D + under Balanced Condition + umbelanced sridqe VedV> 1440 + balanced bridge Ve: Vo, 74-0 vy Veqetaity os 1 me he Wheafstoned cicuit of WheaFsEi nd aidge — 30 A Wheakslines bridge *€ AN Electrical Crud wed to meatun theme Thee Wheat itunes cicuit oF wheartbind ards — 27 “A Wheakstona bridge it an Electrical Cercuit used [> meature the ras aT nce Of Bi liven tn a : Ett cee sy tate ae arranged at Ihe arme “5a: of ‘a quadeitatcral “Aijlveandt of adn fy age & Naf @ onidye baw cen Tenet “Tip +tgkg = 2k -O Aoply kvL fo Lp BCDB +(T)-14) 8 ~(=2+74)x -14 yr “(= mg)a - 989 = (72 +79) 4, fas ust tn valu 94 tau fat Then Lhe bridge is sai anced es Wet Vo 4. From : TT Oe -@ iP ® a Gone eg Fa” ER a f= & [te ralio of sr arm x | “Ove equal) srepuecer rire AL cones SN beagteay ls Connecks « = = — \iemceran F m ‘e . is wee 1m 5 Wee 1 The bride it badantad {8 we |, ” Torso F “ ‘en . =teb ot = - Bet [ie a rr ar Z a : a ¥ a 2013 sel _ under balanced Condition alyanomde shows no dejheebon Cortuit + Find the vate b] nd te rasttonge Ho 9B2 Rp eee ree z Calculate the quivalen resistance between and B ofthe clvrcl network given elo ni) aA bythe ctor, ia bar of HRA. -reitance 1 0 connected acrons the points 4 and B [Ane R =2.0,1= 26] tah from sq, Im quadivtateral AGED! € ye 3 Find the value of the unknown resistance X, inthe following cireuit, if no eutent flows through the section AO. Also ealelate the curent drawn by the circuit fom the battery of -m.£6 V and negligible internal resistance, (Ans. = 6 957 1A] Zag 20 The bridge is balanced ab he 30 2 Xe US 2 ba 3 2 Z aay 2, <7 Ba 6a wae, W_ abe 26a 1 ‘A potential difference of 2V is applied between the points 4 and 8 as shown in the network drawn in the figure. Calculate (i) equivalent resistance of the network across the points A and B, and (ii) the magnitudes of currents flowing in the arms AFCEB and AFDEB. (Ans. R=2.0,0.5 A,0.5 A] a Gol Hatd The onints and & With heme Awland AFDEB. [Ans. R = 2.9, 0.5 A, 0.5. A] c fol Hold The points # aNd & with both hande and’ Pull it out IN fang 20 20 2 ¢ oO E —aa nh Ch azn FD DE = z Tne bridge ¢4 balanced mel ee wo Nee = om i d Currents Mirough AF CEB + AFDER Te current throwgh AFCEB nev aeLa wo ae Wo? “Ty. oon Curent Arough AF DEB Tsyreorbeta Ty 205A Calculate the current drawn from the battery by the network of resistors shown in the [17 "7" figure. rr 2 , M yove resistance + four Tunctone 2 obs Bem 0 ” ‘eb * The bridge is balanced Ter+ 0 e : alae Kw ‘ «Sern es ia te 2 hoo = Rp 2 anh she sae Z a+b oT: M2 aw .2A noe \Gsleulnte the value ofthe current deawn from a5 V batry in the ciceui ax chown. {CBSE (20131 ad Heresistance + 4 Tunctione ’ = 8 3 o/ NeRags 1Sx30 15¢30 4a Aorty ve fo Ame avon + loot, 41554 ~ Gots = 0 25 dom 125 + 3p Aoply kv AT Lrvp 8c D8 lofts ry) -§ (Za4zq]-167y 20 tot, sly sty shy 164-0 10%) 65a = 2034 0 25m - 6m -2 -@ Cavey Kye ts ABcDA ioort sto (2 -24)- ‘at of a ere ferly nut to toop wocerAa joz0 nas Léote +5 [12477] © bor, a sectaay + Tt is a pradical dewice based on fie Principle Of wheat stone brag {ps deerme fe une neu retittance of the qiven wive end pan tke Pungple of Wha aketone bridge bak —4 Ft IE alto called a slide wire bnidge (wmelrebrdge] ts RB» gesiztane box: =e, of, Imetre Long wire Ae Made oF Conitantan or manger’ died along a aicak on @ Wooden base» The ends and c of a wire are !zoined to fisy L~ Shaped Copper dir’ Pa Carrying ganeeh screws - tn between true Copper 4irie4 tnere tu a third ak Carper atrie having these Conne ong screws + The middie id cennedsd Tove Terminal of Be qatvanoeneler and ne othe terminal 0| galvangmeler is Conmctid Bs a shiding Jockey B The Zockey Can be 'made fe Move On fhe Wire HeThe Tockey Can be 'made fe Move ON fie Wire Ae * r + Connect Mhe Crawl danam Intet fey Fe ddledt dtu table Valu of Resistona &” inp + Slide tie "Tockey On the Wire Ac LU We gear null de Heckon te Tg =0 Wit ts Cafled as balancng, Leng measlre She balancing eng th from end alts 3 under bal anced Condition — From 05 @ ee -O toe Foon. 4 \ La Lengin of Be Unknown ratistance 1+ Yadwafdsamdir is found ‘ubing ! Screws gauge ‘- ions baooon th reir in meter bags mde of hick copper sip? fered obtain the balance pointin the mide of the meter bridge Aner Flowing te ” a) 9 thick copper atr’p offers negbigrole resistance 49 teat it dee Tot alter Me value of tuistama wed in the mole bridge 8) Tf tie bakance print v4 taxen in the middle «tt i4 done fs minimise he Percentage error While Cal culating fe nkarion Yaistan a. ©) Generally alloys Uke manganin/constantan] ni chrome are Wed in mefre bridge bechule thane maferials have 04 Teenperaliare Coeff cient Of Yebid bvitg ste tose rn. from the end 4, whan an ae 0) Z] 905 ¥ are, infer oad Vee Wu deflection t x * 3we-7 Yu Xx “g00-3f aad T° aes P51 2 300 Pee (a) Ina meter bridge the balance point is found dis =. to beat 303 om trom the end Ay when the A Sy"™L_8 "PL Settance Yo 128 0s Decne the resistance of X. (@) Determine the balance point of the bridge if Kand ¥ are interchanged 12S 9X 2395125 os 7 Xe 82a tuck Cony sSrine alors meq gots Tasst ane and dou net alley the balan ConditionHs Ges een Tuc Coppen stripe offers neqluqone Yast anu and dow not alter the alana Condl Bon b) Th xs y are, inferchanged = 2s. g2 a wah destection 1 1 > oles (o-d] 262! mast + he bossa Z ©) 11 he Galyanamalec and fhe Col) are indinchanged Sin thy tdesbond of Palance Beint Tremaine Unchanged Ther Javandmdir Shows a Constant de fle chen We Zero def: In a meter bridge shown in the figure, is found to be 40 cm from end A. Ih a res connected in series with R, balance from A. Calculate the value of Rand S. sd cate ME NUMNAEHEEEIR cae v9 te fle ES ge a 23-0 2hsa0 3s. + fom > anzé 420 3s z [CBSE Patna 2015) Bsus 220 uP 20 eee xe Precaations: (0 In this experiment the resistances ofthe copper strips and connecting srews have hen These resistances are called end-resitances. Therefore very smal ‘not be found accurately by metre bridge, The resistance § should not be The eurrent idge wire fora long time, otherwise the wire will become he wuld not its resist incuit diagram ofa Metre Bridge and write the mathematical relation sed to determine je of an unknown resistance. Why cannot such an arrangement he weed for measuring TCHSE East 2016, CBSE 2019 35/41) ‘galvanometer, should be low to ensure Copper strips and conne ‘measured. This resullsin age eetor in measurement. esitance of Sis conneced in paral wl he ll pet cers 50cm rom Determine the values of nd S etl 2 rs om eZ e ae Bo = 3 solving © 40 a(3o+s} = 40 oa ae ane. go : s(Se)0 S9s3 5s pg. sey ca a Calculate the valuc of unknown resistance X and the curent drawn by the circuit, assuming that no current flows thrpgch the galvanometer. Asyme the resistance pet init leapt ofthe wine AB's heat ivem [An X= 601-304) ag at mu dticbon a 26m Hoocm., RAB = o1x100 ae eA een ne 0 a ajoocm, Raw = oxl00 = om Bee ha. eo z 329) lB sv In the m o ‘balance point? 4a ap casei) a ea At mull defitection zx. Jt a t x 1-0 ‘Ze iop-" x st -© (e) 7} fe nol efiedeon eacrnaye! mg of a seth baed galgnced wheattlone Unde Salarcad erection ae 4 T Pal, Bet oot Ree ani” ‘An unlmown resistance Xis now connected in parallel tthe resistance $ and the bie bain a frmmull for Xin term of ly ad's. = ies (CBSE (a 2017} point is found ata distance J 0 deflection Dy cater tad defect rn When "x it Connected ® Paral! — 90h Wik $ Seg = AS. ies mors x cateci AL MUU de Stection Sek B+ de x Fh Creek) Seq (Toords Ss Mol tat) Boe de * xe” toot skpeesye be $F) ae cous vent : + Emg(£)— 0D — | Terminal P2[Vt)— pd — ciaudarust ’ h Emile: Tt is the pote rence Measured across The iene of Mn coll when no cdltent "drawn from the nt aya + Enteral parle (Hi jaar. let yu fe Where “Y'8 Called ax tafevnd raistance of tha Coll 6 due fo Ie ehedtralyhe 5 Chedrede * = + deparation b/w the Ekechrode + "Temperafare (tt. tH) + Concentration 5 nalure of Sfactoly fe eis Ihe Dolenkial dilterence measured across tne 0 ende ob+ mui + seraration vf Ane Bxeutroos + ~ remreradure (TT, 104 + Concenfvation & naluve 0} lectvoty fe Tris dhe Potential difference meaiured across tne Lipo ends of ihe cell When Current is drausn,ftom Ma Cell = ‘The plot ofthe variation of potential difference across combination of three identical cells in series, versus current is as shown here. What isthe emf of each cell? [Delhi 2008] [CBSE (Central) 2016} [HOTS Z equ, Put E=0 oy fem 9 3826 4&4 0aV 3 mie =Vve Shope of fiw V-z Jam Teg = 6a rand = 60. ar we nak Plot a graph showing variation of voltage Vs the current drawn from the cell. How can one get information from this plot about the emf of the cell and its internal resistance? ICBSE (F) 2016) Pry vt The following Gremes snow Lh varcakion of pD-v" Ser ace ind 0 EMF Cy Enkernd valefana. cr} Maumum carat ‘st (emg [open circuit a From qrath B wn re bev wh we Ve €-r1, purze a , Vs € -0x" > E-V.8V 7 oes av tanta tan Ban (3) Internal Passtane 09 z€ -onr > E-Ve8V 3 “Bs 8V reer rr CH) Internal fuistang 9 © Ts =vedlate of Ame Line ye. 8 2 605 Te 28H tH WH) Ts Tmax When & is keost be vinesrY 2 Tan 02 baat Tmax or Te Tmax, wren Ris feast ie Re0, Vere eo Ge From q7aPh, Ve Oy Ime = A cell of ems! and ineenal vacctont x’ is connected across a Warinute trhern Waatonce'k =plot a Graph te thas the Variation the Buti ks ‘cana tote Jol + BS wid # Conencirusit) = ik independent of ‘stent a9 ‘ Wan elation z ie = * mutliPly & on bolas TR ‘anh Aas consider Lion colt of Emits wf eft on. tf ee £1 4 £, having tnfernal Yast Wate Connecky in serves aw Ahown in 4 Ve 8-5. We €2-I% ee tttee tare a warsniy tet * Vax Syorgya fot te Yesmarh dem vnsia = ised 27 terre) » Viplal = Wt Ve dombare With Eto tal on Te ron I} There aren” number of cold Shad eS eee he een = tn os consider fe Cele £4 § €, f =e having internal tesia tances Ty) Connecled in Paralld au shan in 9 Tn pavalld Tas, wv el ean nee vemains same seh ey 7 Sr may ormeg not i ani “i "emah samt rere een Re connate if 4-3) Be ote hoe) Vista « Efi 7 Neier vere) ieesz ‘Te Ta eee Find an expresion for Infernal raistana 1 in terme of e,VaR ce fat Consider a cou of emt € St Anfernal Yaistana “r’ Connecta Be anenternal raistana Rab shown in FY ne wma e bol da TR = ior ve ee @y While me asurin between te mmeatures potenbral dijperence Measured Vous [ vis a errer Wy tacaute tne Volt meer draws Current we should mane a ‘device whi Un measure Lhe Potential ditterence djw Lhe Lwe Poi ithout drawing the current ve ‘nee dehice i fohenhomeke 7 ce tential ox ATH 0} Polential tr weston necwre ; av ‘ > t + S © 2 +TPVW Ve = Tete Pp©s= STEW ee eit A + TEM OW 9 faa, 08 w she 2 217 Mewes fF . VEBV 8 hen a, used Ekrough o Udigerm Crome en Bal vor actost any portion it dire dly trorortional ty th of that Portion "ee ge * Wes BB lne Vac = Klte “Wace @ tac, Vv, &123V, 3.220 + Sdriver > Sunknown + Re wihad Uke o o kafuly dec for th ill omelia tne following is tne Corwit duagram 401 ‘Comparing dhe emf of Mie cenle “insert Key'k’ 4 adzust tne P Rreostat “en 40 that a shady current 4 pamed twrough the Polenkiomerer AB + insert Ka 4 Temove kr MoVe The Jockey on the wite AB + Insert Kr 4 Temove k; ee + me ie sone on te wre A LU 3928 ap all pecton as TR Vea = Vaz a - Séee [Hf] v8 fh oo ota + RD oe] age elk -© Coma) aa -a-8© GEIRNEO (2 Rezt] Motes tnleenal restance, 7 (E 1] With The hdP 0} 6 Cixtuit diagram find an express: of The Cell Using Po ntomeler pe allowing ene Gpaut diagram |: r Mme aburi inka nal faistana | the call ung rokendbomde 7 + stepe of working ee aes Adgust Mis Rheosbal Rh 40 that a shady current i4 patted through the Poknbomdler Wire AB . — key ey ond Temove Ka + insert + move the Foley on ee wireAs Ki tg:0 Rove hel sauce Toceey on fhe Wire AB Tq=0 1 At wu de flection Vea = Vata Ves Ze “Mes E Ste a “Ve + FES |e Ve = eh pixezd) serene x ————e + Sensitivity of o Wheatalonial/ mete bridge + fllsne dour resistances should have Neary the same Value + mieoridge, the bal auld be r waprint’ of Be mek nage 6 ggg (should be Anau) +BY increading the Lengik of the polntiomel wire Vise ay east at BY Connecting a high Tedidtance box in the Primary Giruut Then TY, WK = 42S Aenm BW "of Lhe Palentiomefer Can be incteated Dowie _Iiwa——««€ | /* 2) im the graph shown below for two potentiometers, state I son which ofthe te potentiometer A oF B, is more 4a (i) SORBLAWIY » Me Vadus Of K Should be a Small Value Wnich Meand fw pelenGomeler Can meaaure ¢Ven fhe Amablat Value 0} PED 8994 Specating Jog Oh akin omdersie (9, BY connecting o hrevist anc “box in Pokntiomeler & 14 more Seria te the Primary densifive than yim, Behe eeABS KA Areist ance “box In Pelentiomeler & 14 more Aerta 28 the Primary Sensitive: Fann “Wh ong AL a jasv, a5em th 7 a 68am fa. 1:25%63 .2a5y 3 llatemt Var shown inthe igure. When ted wa diver wv [ape in the secondary circuit the balance pot sound ‘eb om. On replacing this cell and wsing cell of wk he balance pein she to 80 (@) Caealtesninown emf of the ell . (i pain wth eon, whether the cet works the diver om) celisreplced wihacelloanf 1. (ii) Does the high resistance, wed n the secondary circuit afc he halance point? Juntify {CBSE Dat 2008} a0 et Las Than Cel in the Secondary Cirewt because the fetal vottage acres The #8 ce IV land hence Xe balance point “Cannot be obtained or Atoaye Edriyer > 2unknown a eo, aa ®’ Will Oat Like a safety device for qalvano mba Ina potentiomet in rangement for determining the em ofa cell, the balance point of the cell "When a resistance of 9 £2 is used in the external circuit ofthe cell {he balance point shifts to 300 cm, Determine the interna resistance af the El I y= 3seum, Reda te fle tle ts S00um , 127 t [x ) rr (Bl Tt + sBetsa ] Yetsa ithe potentiometer circuit shown, the nul points atX. State wih eason, point will be shied whens * {a) resistance Ris increased, Keeping all other parameters [~ 21 wmchanged (©) resistance $ is increased, heeping R constant. e [CHSE Bhuboneshwer 2015} Let 1 be the balance length of the segment AX on the potentiometer wite for given resistance Manel S as + balanang tengit widk increate and hena it with “shite towards B 6) Ey st, ropa, Se Os it Wi ak tke o sapedy devia “for the galvanomeder ye balance TCHSE Patna 2015, fd refer above {st Xa Y ver npr on pen pert cig he si [gente he Wop shor hinge irre ae Srne(ay en Th mete oh (054 RY Th betes wee « beban dng Length will increate 6, hence st He Sante piles If Sb, xt We ee Et 247 constant 2 MT increase, V Will decrease Bo ly = badancng Langit will decrease and hence Wik AW tt Gloarae A’ or Wy = ERY (Keeping tre Current constant] Ve will decreata OY uit decreas [ vast) 2 balancing fength Will decrease and hence WikaW te Doloavae A’ "Sve students “07 and perform an experiment on potentiometer separstely using the circu given- Keeping other parameters unchanged, how will the position of the mull point be affected if 1 (0 5 increnes the value of rexiince cpg the Resse and (i -7 decreases the vale of resistance Sin he sey the hey Kyran open andthe hey Ky hse sis the setup by youranwerinenchewe. (CBSE (F} 2012] HOTS} oh (0 Then re, be bre 7 “$= WV. + balanung Jeng will increate and henc it tr | Wil Shih Solyards ©. UO TE SY, Re id open Ce ye Cir Gut is open Circuit 2 There te Me Change iM balancing Lengit ‘Two primary cells of emfs wire ¢, and ) are t+ connected to a potentiometer wire AB as shown in fig. Ifthe balancing lengths for the two combinations of the 25 cells are 250 em and 400 em, find the ratio of c, and cy. a by-ta: Kh cosectn &, L a80 avec) ae Kh C9 By By kha + ey = 1x250-0 Eines kaon ME ~ @-O, irés 2 wh» 5 (+80) ~ 88-82 Zit: 3K pei eate y A Be, 1382 Me. as BORUETES) + TWo cet 0] Smt ) areconnecid a4 4rown in hg When a Pakentiomeler it Conneded b/d ASB, dhe balanung feng is found fo be 3m. on Connecting the dame potendiomelir bw Aand ¢. Mi balancing length is af im Calaudab Lu ralio of & 4 22 s Af tot Cased bg ASB Cate Ch) bw ASE ge kh Ee Ki yo ke -O fey ID -07@ 2, = BK w &) + 32)-3hs res TE 2s) = 322 Hi. 2 2 Sees sieshown, AB is a uniform wire of ‘acest auagea 16 Gand fength tm and ‘and The balance negligible internal resistance Rg URNRERAF of is found at 30.cm from end A. Calculate the value of the resistance R. [CBSE Chennai 2015] Rie «160, tae Yap = X21 Vag 25 10V Maa 380 = fax Mn fan = 30cm ee * Nag > T RA8 -O ’ 42 primaryoruit t-2 -® 7 s, Bris 2 * He 8 xine. 2 xs RMS 4 OOSR+ 075 = 6 cast Rris}= 6 005K = 6-945 ays {A potentiometer wire of engi Im has resistance f 100. 1 x comecte to» 6V battery in gy tbe 200 tab tno tm. Raw = toe, tag: noom Wp ge Van =F tan ce — Vag = & tar a|_ va _a . Van = daw Lf jena Vas tng. Vax tag alas Yao = Za x Raw Ta oie: 6 = 6 % gan Van = Spe? muy wy In the following potentiometer circuit AB wire of length 1 m and resistance 100. Cal potential gradient along the wire and balance length AO. [CBSE Dethi 2016] {HOTS} ol k= Vaw = Tapx Rag ry Ta but Rap: ton , tag=im Vee = Trax 2a 2 x50 S isorto © Vag s 2x50 J aah # tart, Vaz = g90mv = 0.3 Raw: son, tags tommo ~ ves gooey 1 Naz » 300m) = 0:3 Soa, taps tom 2K dap s0750 $00 100 x50 2 0sv ymple 3.10 A resistance of Ré draws current from a potentiometer ‘The potentiometer has a total Fe : 5.29). A vollage V is supplied to the potentiometer (DEFER te YAN wn ang ona ne ‘the potentiometer. ty ‘al 1 . Loe eu a eT a a es ins aaa * bac = Hip HRs? “aT Rest8) Rott = Rac rey 2VCRortR) x RRO Rae > Rap + Ree 7 Teroree — feerety Efe + bo Bev ek Taig sa Bye» ARO + Ro (Ro s2a] eee) =e Bits Tea) T= av[Rorae) Rae = 2h ee tte RRO ta Vabe TERo+e8) Drijt vdocity ( V mtn aay With Whe ch an e° 64 drifted under the Angluenice of the APped Stearic HAD Va = Vayy = heer tn 0 + Relaxation Lime [ t) Hrespath:- Tt 14 the Lengt where coumsion doce not take Alaa b ‘us 4 Dn a x fer the electron v4 relaxing before colb ding With te next alom THe Emme ie Called Ydaraban fine‘ ¥ fiere the electron is thawing beSere Coli ding Ait te next alom Tut time ie Called tdarabon Lime tn “4 tte 7 1] Semperabure Ti and KE Tren Lie Trereased, fhem The Ydaralion me ao ‘bay finan carreavon for anit vavoty ce VaREC/ MER Contider a. Conductor of Length t 4 Saif tnacboa ede 4 Shown Sh er a Ew anpte » Use stn = 0 -O fi applied Sey» Ha pont _Prlgmnnennor ag nen a po of V Volt is apphed fren tie letra gain éne rg and Glide wite fie adome and foee AW eafea Energy there by it Trave? wut a nit ed iratin Se Moy? tae ‘Va 40-@ ie hear wa Tn + iy ote, . Uy 2 des Bs Giardina ti eae) a i ig ne a> [Das so] va tty oot) Hes sta) of Lette wut Te feetpensta . ET | umber dened’ is Cqmal number A Obst Predent in unit Volume consider a conductor 0} Jengmi 1° 5 area. 0} cross section “A” hawing Mumber df e°4 Per Ur Volume Let oD 0} V Val 1s Connected across the conductey n— Im? SE t “4 1A Seeagned al eT Of Current enkering normal nM fartonduclor Wperunit+a. mu Find Ine Ctpreshon ie Mhe " of tdax alron time Consider a Conductor oy Jength V's area 0] C1044 Section fh’ connected ty a ballery ft Vovolt then tke Current “r' 18 given’ B= neava —O (derive) > et -@ stance of a Conductir i'n terms a_-+t bur smokey = x oF iM on or BE Coma 0} Cross Section ‘obec by a baltery of ¥ vos PA ‘constant? TEBSE 2019 551))- ¢ “ OR dame a lo ok: Saw Pe cncanrnana arene nseres eee ee ee ee eras tense anon jh dE = NeaW 71 oo Tn Aerie qe ty fog Way = My Vay CW? NERY, we Myvard og ¥ NEN, = eM Yay se a [ia Sa a Ray g es a diteence Vi appli acrons the ends of copper wire of length he eflect on deft velocity of electrons if ) Vishalved? (i) Fs doubt i) ishatvea? oat [pea jou 4 AR) ekyested wh od cal £, re ). cis doubled (Qh gga) eae [Se[ey z <1 3 = ers ore ne Gncrease in Tusk bvily eer Yaiebvly. fer wat ee Rfe-&) + Semicondu efor ( Ge of $i] de -Ve, Tf tre kemperalire ig increcaned “it tae tana raatvily (will deerme + of Amal posi tive Vauus + Ty Aemperabaire 1s in Creaved hen Tewstaie | ves idte vidy WIM incre abe + mel C cere) - ibe Vi ny hemp tats iMtreatea then TuistanG/ Tuishevity tu need tein femper alte Sumit ft olor ot z— Condy cre which do not obty onmé aus ore wy omic cond x Gemicondu cher makervale, t Sy (ii) Non-ohmie i | SS 3 Po -AOns 2 ao t Non ohmc 3] behaviour Tailors f celour code [carbn testtor hat four Coloured bande Wich Tepresenks Ihe Yai stane an wd rary wba, Toherane error oot 234 5 Tolerant /ervot 2 none — + 20] + dibver — x (0h * Gold = 2 S/- A carbon resistor is shown in the figure. Using colour code, write the value of the resistance. (CBSE 2019 (55/3/1)| 4 videt 2 Green Red (R: gant zany SS fa A carbon resistor is marked in colour bands of red, black, orange and silver. What is the resistance and tolerance value ofthe resistor? = 2. lef tl R= 2OxWAs Io, (@) You are requived to select a carbon resistor of resistance 47 KO 4 10% from a large callection. What should be the sequence of color bend (0) Write the characteristics of manganin which make it suitable for making standard resistance. [CBSE (F) 2017) wwe wae etis passed through them. Which wite gets heed up more? Jusly your answex TOBSE (a1 2017) at FB ws nt manga nd hth copper ecu agi an qual resistance ‘Which ome of these wires wil be thicker? TEBSE (Al) 2012, (South) 2076} LOTS] BER EE Souen > Serr Amanganin 7 Acoptn San ra inarleawe ee i) Resto Sls constant Hee ’ Meth let . 1, > xa [From graph) make mare the i mpaaliive more the vesia Tana “Tet