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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— PI3V3=3xlE'§4_w) M a) .1: It. ‘r: egﬁﬂm +0 : Fm: , 3.0 Rﬂ, amok 611 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. = 3020: [0V :1: V R’ 6 ' _ﬂ (0" END4:401" 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}? AtOV; 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. VTIJR'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+Ea>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 prodons 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) {Zapatawags 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 (WWQmTﬁﬁﬂa)
(6) 03= %= 5/76
(f2 £3= “$0314.: = :fwwi =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: furMM}; (5,2 __—. 9+5; =5/(F
{758:} CI: ENE , Sﬁ
l —"L":'.—' .__.L—_._._
595 57246.5 ' éﬂF*ZLxT=" =35}, @333/F 860’: .51 V! _'= 3ﬂFx'5t/ = 4.5314“
. fore ﬁe .S'aurtc'é' 54 ((5526? . Mm: was no Marge. Oil 61,: mmﬁf l 8 g ...
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 Spring '10
 GRUNER
 Physics

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