Chapter 28 IQ Questions - 28.2.1 EMF and internal...

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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‘-rR-2r‘=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 2-i2R2- 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 VA-VD= a. Taking the path DBCA gives VA—VD : b. Ans:1. Extra: When R1: R2, VA-VB=VB-Vc= 2/2. So VA—VD : a : i2R2 — £2: sf2 — a = — 9’2. Check: 1: = — ez-i1R1+ £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 RC-circuit 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 VB-VA = — i1R= — 8 f3. Ans=3. Extra: Consider the branch ADF. Since R3: R4, VA-VD=VD-VP= 8/2. Knowing the potential differences between AB and between AD, one can find VB-VD, 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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Chapter 28 IQ Questions - 28.2.1 EMF and internal...

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