Sample Midterm 2 Solutions 12-18

Sample Midterm 2 Solutions 12-18 - '51 Saw? a" '...

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Unformatted text preview: '51 Saw? a" ' .. 13:1 R€§"§+R‘EO“*6”@ PM (:75 J,R‘5=E=E . on)?" PD 1‘5": BM‘FPE C ' 1} R2=33x6=1r w m— P-I3V3=3xlE'-§4_w) M a) .1: It. ‘r: egfiflm +0 : Fm: , 3.0 -Rfl, amok 6-11 m {m [MIME M 91:. Pvmaf Amp by R w V.=I.R.=4~<T:2°v Iii -‘ P0 I ‘ u : Ti}: E 15‘ V3 -‘ («E—V. = 30-20:- [0V :1: V R’ 6 ' _fl (0" END-4:401" W (shown above) I E _ E] (a Two batteries. three registers and one switch are connected as shown below. Initially the switch is open. Asemne that the potential at pointA is 0V. :1} What is the potential at point E? (Vim my mum b} Whatisthepntentialalpointfl‘? - ~—I‘—I—-—I__-—-———'_'—-_.I_._ e) What is the eqmvelent renistmtee R. of the circuit? d) What is the relation between I; and 13'? What are their values? e) What is the potential at point D? f) How much power is consumed by resistor R}? At-OV; F E b) V 3 20V (No if} o‘i’mi‘t'ni drep Cures: RL1:R1+R3=(3+1)£L: R”) Lila: Eszovatm gel 541. V-TIJR'JT-‘i’l: 3V Ptiieva "7 ‘-' '32. W Now the switch is dead. g) By applying thejunetion theorem ofKirehhofPs rules at the point C, showthe relation between the current 1.. I; and I3. 11) Next by applying the junction theorem at the point F. show the relation between the eurrent 1., I; and I; again. Compare it with the answer above. Do we really need this equation? i) By applying the loop theorem of Kirchhoff: rules along the loop A—+B—+D—rE—>A, show the relation behveen the amt I. and current I}, j) Finally by applying the loop theorem of Kirchhost rules along the loop F-+C—+D-+E-a>F, show the relation between the current I; and current !3_ 1:) By combining your solution: in the previnus page. solve II, I; and 1'3. I) What is the potential Vat the point D? (Mum: that thcroimd has 01’.) m) How much power is supplied by each ham, 5‘; and a}. n) Show that energy is consumed in this circuit. Lot‘s explore the pro-dons problom in a dill-'th way witlmut applying Kirohhofl": ruins. Aotmlly tho aivon circuit is equivalont to the circuit below. Hue, we are allowed to combine two any“: in parallel and replace them byjustone Wwith 3V, bocauao tho potmtiul at pointderisalways 3 Vanyway. Calculate [ho following; you should get the mo mm: as tho preview page. on) Botfivaleutreaistmoo R". p) Cmnt 4'3. :1) Potential Vat Point I). (Mammoth: the pointxl has OV.) E a) b) c) d) 6} Now We move on the circuit with Capacitors shown below. Initially the switch is open, and the Capacitor C; is miehmged (Le. Q; - 0), but Q; and Q; are already l-‘uullz.P charged up by the battery 3:. Calculate the following. The omrent I (from point C to D). Equivalent capacitance C". Total charge 9.... stored in C... Potential at the point D. {Assume that the point A has 01’.) Charge Q: which is stated in the capacitor Cg. (51) (UAW M, (V) l)- Energy E3 stored in the capacitor C3. 01 £1=3V (fl) Meow I =0 Cf=3pF 52:31! (la) {Zapata-wags CWQCB m Msarlké Vfié‘K’fi“ 9&3 =32"; 2;; =ffii‘p, 653:2/4' (‘5') GM —C¢5.83=-2flp fly = 6/5 (d)1/2>= 95:: {fig = N (WWQm-Tfififla) (6) 03= %= 5/76 (f2 £3= “$0314.: --= :f-w-wi =3flJ In the: same circuit as shown in tho previous page. the switch is now olosod. Calculate the following @ quantifies. (Hint: You can play the sum: trick as the previous Resistor pmblom in Pop 6. Connect Ihe pointX and Jr" by an imaginary oonducting win as shown bclow.) Calculate the following. 3} Equivalent capacitance C". 1:) Total charge QM storod in C", i} Potential at the point B. (Ame that the point A has 0 V.) j] Charge Q, which is stored in the capacitor C3. 1:) Charge :1 which goes through the switch (from point B to C} just one: it in closed. D A {=ow F’ E (3) Capacmrs (3; & C; are or: fur-MM}; (5,2 _-_—. 9+5; =5/(F {758:} CI: ENE , Sfi l —"L":'.—-' .__.L—_._._ 595 57246.5 ' éflF*ZLxT=-" =35}, @333/F 860’: .51 V! _'= 3flF-x'5t/ -= 4.5314“ . fore fie .S'aurtc'é' 54 ((5526? . Mm: was no Marge. Oil 61,-: mmfif l 8 g ...
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Sample Midterm 2 Solutions 12-18 - '51 Saw? a&amp;quot; '...

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