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Unformatted text preview: 28.2.1 EMF and internal resistance Given battery has emf: 8 =10V and the internal
resistance 1:152. Consider two cases: Case I: R=0.01 9.
Case II: R2100 $2. Compare V AB with 8 for case I. 1 2
case I VAB<< 8 VABN 8 Hint: Verify that viii—ERR
" r+ Extra: For case II, show that VAB ~ 8. Explanation: Simple calculation shows for case I:
V AB=0.1V<< E . In other words, when r>> R, most of the potential drop is across 1'. Extra: For case 11, VAB=9 V ~ 8 . This implies when
R>>r, most of the potential drop is across R. Combining
the two cases together, it implies that the most of the
potential drop is across the dominant resistance. 28.2.2 Equivalent diagram Consider resistor network below. Determine RAB. 1 3
iRAB 7R/6 6R/5 5R Eguivalence rule: A segment in a circuit may be
contracted to a point. A point may be stretched out to a
segment, Explanation: Apply the equivalence rule, we obtain R 2R
A—l’w’lil—W’l—B .._JW,— —"/\/‘I—
R R RAG: RC3 and RAB AIlS = 282.4 An infinite chain
A r A,E r
———Ml waTav
Ear < 2r 2f
B r5 < E. + B’ E
Given an inﬁnite chain with a repetitive pattern as shown. One may determine R (R AH) by ﬁrst writing down an
equation for R. The equation is given by (choose one): 1 2rR
r +—
2r +R
2 r+i+l
2r
3 R=r+21+R Hint:
By inspection, the resitance to the right of A’B’ is the
same as R or R. So the iniﬁnite chain can be redrawn we as follows:
) \
[2 —\/\/\ Explanation: From the hint,
1 2 IR RAE =r+:£=r+ 2r +R =R
2r R
Ans: 1.
Digression: The above equation leads to R‘rR2r‘=0, or
R=2r. 28.2.3 An equivalent network 0 Given a. current I enters the resistor network.
0 Find: i1. 1 2 3
i If2 [.33 U4 Hint: Steps.
Find RAB, find VAB, and then i1=VAB/R Explanation: Based on the equivalent diagram given
below, RAB=R/4, VAB=IR/4. In turn i1=VABIR=I/4. Ans=3.
R ITi
A_, 1 R B Rl2 28.2.5 Series and Parallel Circuits 3 identical bulbs are connected in two ways as shown.
Denote I and P as the current and the power in case 1 and
I’ and P’ are corresponding quantities in case 2. The ratio
of currents through the bulb B is given by: 1 2 3 4
133/1B 4/3 1 3/4 2/3 Hint: Redraw the diagram to display more clearly the
series and parallel connections. Extra: Show that the ratio of the powers: P’ A/P A=9l4 Extra. PA_(I'A)2R [PB ]2_[3 PA (IA)2R— 113/2 2 28.2.6 Series and Parallel Circuits case 2 3 identical bulbs are connected in two ways as shown. The
batteries of the two cases are the same. Denote A and B to
be the brightnesses of bulbs A and B in case 1, and A’ and
B’ to be the respective brightnesses in case 2. Compare B
with B’. Hint: Redraw the diagram to display more clearly the
series and parallel connections.
Extra: Compare A with A’. Explanation: Denote I and P as the current and power in
case 1 and I’ and P’ are corresponding quantities in case 2. __ _ _ 3
IB _ V/(3R/2) _ 4 (I'A)2R _[1'_B
PA ‘ (1m: ‘ 18/2 . But “lezR”, so B’<B. 28 .3 . 1 Kirchhoff rules Hint: For loop ABCDA 8 2i2R2 8 1+i1R1=0.
You may multiply the equation by —1.
Explanation: From initial point to final point rules are: i— f => Vf—Vi=+s
Rule 2) R I
I4VV— f :> Vf—Vi=— IR By applying these rules and then multiply by —1 one
arrives at Choice 3. 28.3.3 Two loop network Current I enters A and leaves B. Given: R1=R4=R5=r, R2=R3=2r. Determine the ratio i /il. l  1 f 2  3 I
Hi 1/2 2/3 1 Hint: From symmetry, i1:i4. Then i5:i4—i2:i1—i2. Write
down the loop equation for loop ACDA. Explanation: The loop equation is —i1r—(i1—i2)r+i2R:0, or
i2r+i2R22i1r. So i2/ i1:2rl(r+R):2I’3. Ans:2. 28.3.2 A cubic network Given a cubic network of identical resistors, each with
resistance r. The current eneters the network at A and leaves at D. Find 11 and 12. cz MN D I w
Ir“ .1 v
>
? A B‘/_IM
g I V >>>>>>>> 11 Hint: Make use of the symmetry property of the network.
Extra: Show R AB=(5/6) r Explanation: By symmetry, at A, Iis equally divided
into 3 branches. So Il=I/3. At B, 11 is equally divided between two branches. So 2=Il/Z=I/6. Ans =4.
Extra: RAB=VABII=I (1/3 +1/6 + 1/3) r/ I = (5/6) r 28.3 .4 Two—loop network Consider general 2—loop network. Current flows from A
to B. Loop ACDA gives —i1R1—i5R5+i2R2:0. Determine the equation for the loop CDBC in terms of i1,
i andi . Hint: Use i1:i3+i5 and i4:i2+i5.
Extra: By inpection, would you be able to find il in
terms 1? Explain your reasoning. Explanation: The loop equation for the loop CBDC
gives: i3R3+i4R4 + i5R5 = 0. Substituting the relations
given in the hint leads to Ans:3. Extra: The two loop equations of ACDA and CDBC
together with I: i1+i2 give three equations for the three
unknowns i1, i2 and 15, so i1 may be solved. 28.3.5 Potential difference betweem 2 terminals Consider the setup shown. Define following two potential
differences: 0 a: i2R2 — 82
o b = — £2 i1R1+ a, 4 both are
incorrect Extra: When R1: R2, £2: £1: E, show VA—VD =— 8/2. Explanation: Notice that i320. Why? Now we apply the
Kirchhoff ’s rules. Taking the path DBFA gives VAVD= a.
Taking the path DBCA gives VA—VD : b. Ans:1. Extra: When R1: R2, VAVB=VBVc= 2/2.
So VA—VD : a : i2R2 — £2: sf2 — a = — 9’2.
Check: 1: = — ezi1R1+ £1: — e — £/2+ a: —£/2. 28.4.2 RC—circuit: Charging
[“'W""”:1:q
C "q a l
l' R S \t Consider the RC circuit shown. The loop equation is 4
given by E—E—iR =0 . Determine q and after closing S at t=0. “in ' immediawa Extra: After a long time (i.e. at t=°°) there is no more
current, show that qm=£C. Explanation: The present case corresponds to charging a
capacitor. There is no charge initially, ie. at t=0, q=q0=0.
From the loop equation: izigzsr’R. Extra: At t=°°, no more current ﬂows, or i.°=0. Loop
equation implies that qzqwzsc. 28.4.1 RCcircuit
r C1 C2 Egan Given: C1=C2=C. Close S at t=0. Find i at t:0. iat t=0 Extra: Show the plate charge on C1 at t=°° is sC/2. Explanation: Consider the equivalent circuit.
R=r+ r/2: 3r/2. C12: C/2. At t=0, no charge is on C12,
Vlzzqt’Clzzo. i: s /R=2 3 /(3r). Extra: At t=°°, 1:0, plate charge q: a C12: 8 C/2. 28.4.3 Steady state currents in presence of C The circuit is in a steady state.
Given R1: R3: R4:R, R222R. Determine VB—VA. VB'VA 2/3 ‘ 2913 Hint: At a steady state, by definition the charge on the
capacitor is constant, so the current from B to D, i=0. Extra: Show the potential difference VB—VD: 0’6. Explanation: Since i=0, the potential difference across
ABF: 3. i1: 8 ."(R1+R2): 8 /(3R). So the potential across
AB is VBVA = — i1R= — 8 f3. Ans=3. Extra: Consider the branch ADF. Since R3: R4,
VAVD=VDVP= 8/2. Knowing the potential differences
between AB and between AD, one can find VBVD, i.e.
VB—VD : (VA—VD )—(VA—VB): (9’2) —(t»:f3) : 8/6. 28.4.4 Characteristic times of RC circuits A Fig. 1 Fig. 2 Consider the RC circuit in Fig. 1. Given R1: R2: R3=R4=r. The capacitor is charged initially
to Q0. Close the switch at t:0. After the switch is closed,
at any t the charge Q, on the capacitor will be given by Q=Qoexp(t}1). Determine this characteristic time 1:. Hint: Convert the present circuit to a simple RC circuit.
Extra: Show that if we “short” R2, i.e. connect a wire across R2 (see Fig. 2), then 1=(2/3)rC. Explanation: By inspection, the resultant resistance R
corresponds to having two 2r—resistances connected in parallel, so R:(2r)(2r)f(2r+2r):r. Or ‘c:rC. Extra: For the new situation of Fig. 2, by inspection,
R:(2r)r.n"(2r+r):(2l3)r. Or 1::(2f3)rC. ...
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 Fall '08
 Turner
 Physics

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