A positive value is obtained, so the current is counterclockwise around the circuit.

Chapter 7. (a) Let i be the current in the circuit and take it to be positive if it is to the left in. We use Kirchhoff s loop rule: ε i i ε 0. We solve for i: i ε ε + 6. 0 050.. 4.0Ω+ 80. Ω positive value is obtained, so the current is counterclockwise around the circuit. If i is the current in a resistor, then the power dissipated by that resistor is given by P i. (b) For, P i (0.50 ) (4.0 Ω).0 W, (c) and for, P i (0.50 ) (8.0 Ω).0 W. If i is the current in a battery with emf ε, then the battery supplies energy at the rate P iε provided the current and emf are in the same direction. The battery absorbs energy at the rate P iε if the current and emf are in opposite directions. (d) For ε, P iε (0.50 )( ) 6.0 W (e) and for ε, P iε (0.50 )(6.0 ) 3.0 W. (f) In battery the current is in the same direction as the emf. Therefore, this battery supplies energy to the circuit; the battery is discharging. (g) The current in battery is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging.. The current in the circuit is i (50 50 )/(3.0 Ω +.0 Ω) 0. So from Q + 50 (.0 Ω)i P, we get Q 00 + (.0 Ω)(0 ) 50 0. 3. (a) The potential difference is ε + ir + (50 )(0.040 Ω) 4. (b) P i r (50 ) (0.040 Ω).0 0 W. 059

060 CHPTE 7 (c) P' i (50 )( ) 6.0 0 W. (d) In this case ε ir (50 )(0.040 Ω) 0. (e) P r i r (50 ) (0.040 Ω).0 0 W. 4. (a) The loop rule leads to a voltage-drop across resistor 3 equal to 5.0 (since the total drop along the upper branch must be ). The current there is consequently i (5.0 )/(00 Ω) 5 m. Then the resistance of resistor must be (.0 )/i 80 Ω. (b) esistor has the same voltage-drop as resistor 3; its resistance is 00 Ω. 5. The chemical energy of the battery is reduced by ΔE qε, where q is the charge that passes through in time Δt 6.0 min, and ε is the emf of the battery. If i is the current, then q i Δt and ΔE iε Δt (5.0 )(6.0 ) (6.0 min) (60 s/min). 0 4 J. We note the conversion of time from minutes to seconds. 6. (a) The cost is (00 W 8.0 h/.0 W h) ($0.80) $3. 0. (b) The cost is (00 W 8.0 h/0 3 W h) ($0.06) $0.048 4.8 cents. 7. (a) The energy transferred is U t Pt ε r + ( 0. ) ( 0. min)( 60s / min) 80 J..0Ω+ 50. Ω (b) The amount of thermal energy generated is U i t F F HG ε I HG + K J r t 0..0Ω+ 50. Ω I (. Ω )(. min)( KJ 50 0 60 s/min) 67 J. (c) The difference between U and U', which is equal to 3 J, is the thermal energy that is generated in the battery due to its internal resistance. 8. If P is the rate at which the battery delivers energy and Δt is the time, then ΔE P Δt is the energy delivered in time Δt. If q is the charge that passes through the battery in time Δt and ε is the emf of the battery, then ΔE qε. Equating the two expressions for ΔE and solving for Δt, we obtain

063 (b) The simultaneous solution also gives r 0.30 Ω. 5. Let the emf be. Then i i'( + '), where i 5.0, i' 4.0, and '.0 Ω. We solve for : i (4.0 )(.0 Ω) 8.0 Ω. i i 5.0 4.0 6. (a) Let the emf of the solar cell be ε and the output voltage be. Thus, ir F H G I ε ε K J r for both cases. Numerically, we get We solve for ε and r. 0.0 ε (0.0 /500 Ω)r 0.5 ε (0.5 /000 Ω)r. (a) r.0 0 3 Ω. (b) ε 0.30. (c) The efficiency is P / 0.5 received 3 ( 000Ω ) ( 5.0cm ) (.0 0 W/cm ) 3.3 0 0.3%. 7. To be as general as possible, we refer to the individual emfs as ε and ε and wait until the latter steps to equate them (ε ε ε). The batteries are placed in series in such a way that their voltages add; that is, they do not oppose each other. The total resistance in the circuit is therefore total + r + r (where the problem tells us r > r ), and the net emf in the circuit is ε + ε. Since battery has the higher internal resistance, it is the one capable of having a zero terminal voltage, as the computation in part (a) shows. (a) The current in the circuit is ε + ε i, r + r + and the requirement of zero terminal voltage leads to ε ir, or r r (.0 )(0.06 ) (.0 )(0.0 ) ε ε Ω Ω ε.0 0.0040 Ω.

064 CHPTE 7 Note that r r when we set ε ε. (b) s mentioned above, this occurs in battery. 8. The currents i, i and i 3 are obtained from Eqs. 7-8 through 7-0: ε ( + ) ε (4.0)(0 Ω + 5.0 Ω) (.0 )(5.0 Ω) 3 3 + 3 + 3 (0 Ω)(0 Ω ) + (0 Ω)(5.0 Ω ) + (0 Ω)(5.0 Ω) i 0.75, ε ε ( + ) (4.0 )(5.0 Ω) (.0 )(0Ω+ 5.0 Ω) 3 + 3 + 3 (0 Ω)(0 Ω ) + (0 Ω)(5.0 Ω ) + (0 Ω)(5.0 Ω) i 0.05, i i i 0.05 0.75 0.50. 3 d c can now be calculated by taking various paths. Two examples: from d i c we get d c i (0.050 ) (0 Ω) +0.5 ; from d + i 3 3 + ε c we get d c i 3 3 ε ( 0.50 ) (5.0 Ω).0 +0.5. 9. (a) Since eq <, the two resistors (.0 Ω and x ) must be connected in parallel: eq 300. Ω x + x x. 0Ω. 0Ω + We solve for x : x eq /( eq ) (3.00 Ω)(.0 Ω)/(.0 Ω 3.00 Ω) 4.00 Ω. (b) s stated above, the resistors must be connected in parallel. 0. Let the resistances of the two resistors be and, with <. From the statements of the problem, we have /( + ) 3.0 Ω and + 6 Ω. So and must be 4.0 Ω and Ω, respectively. (a) The smaller resistance is 4.0 Ω. (b) The larger resistance is Ω. b g x.

065. The potential difference across each resistor is 5.0. Since the resistors are identical, the current in each one is i / (5.0 )/(8.0 Ω).39. The total current through the battery is then i total 4(.39 ) 5.56. One might alternatively use the idea of equivalent resistance; for four identical resistors in parallel the equivalent resistance is given by 4. eq When a potential difference of 5.0 is applied to the equivalent resistor, the current through it is the same as the total current through the four resistors in parallel. Thus i total / eq 4/ 4(5.0 )/(8.0 Ω) 5.56.. (a) eq (FH) (0.0 Ω)(0.0 Ω)(5.00 Ω)/[(0.0 Ω)(0.0 Ω) + (0.0 Ω)(5.00 Ω)].50 Ω. (b) eq (FG) (5.00 Ω) /( + 5.00 Ω), where 5.00 Ω + (5.00 Ω)(0.0 Ω)/(5.00 Ω + 0.0 Ω) 8.33 Ω. So eq (FG) (5.00 Ω)(8.33 Ω)/(5.00 Ω + 8.33 Ω) 3.3 Ω. 3. Let i be the current in and take it to be positive if it is to the right. Let i be the current in and take it to be positive if it is upward. (a) When the loop rule is applied to the lower loop, the result is The equation yields i ε i 0. ε 50. 0050.. 00Ω (b) When it is applied to the upper loop, the result is The equation gives ε ε ε i. 3 0 i ε ε ε 6.0 5.0 4.0 3 50Ω 0.060, or i 0.060. The negative sign indicates that the current in is actually downward. (c) If b is the potential at point b, then the potential at point a is a b + ε 3 + ε, so

066 CHPTE 7 a b ε 3 + ε 4.0 + 5.0 9.0. 4. We note that two resistors in parallel, and, are equivalent to +. + This situation consists of a parallel pair that are then in series with a single 3.50 Ω resistor. Thus, the situation has an equivalent resistance of (4.00 Ω)(4.00 Ω) eq 3 +.50Ω+ 4.50 Ω. 4.00Ω+ 4.00Ω 5. Let r be the resistance of each of the narrow wires. Since they are in parallel the resistance of the composite is given by 9, r or r/9. Now r 4 ρ / π d and 4 ρ / π D, where ρ is the resistivity of copper. Note that πd /4 was used for the cross-sectional area of a single wire, and a similar expression was used for the cross-sectional area of the thick wire. Since the single thick wire is to have the same resistance as the composite, 4ρ 4ρ D 3 d. πd 9πd 6. The part of 0 connected in parallel with is given by 0 x/l, where L 0 cm. The voltage difference across is then ε'/ eq, where ' /( + ) and Thus, eq 0 ( x/l) + '. P ε ( + ) ( ) ( ) ( ε ) 00 x 0 0 x L 00 0x x + + + ( 0 ), where x is measured in cm. 7. Since the potential differences across the two paths are the same, ( for the left path, and for the right path), we have i i, where i i+ i 5000. With ρl/ (see Eq. 6-6), the above equation can be rewritten as

067 id ih ( / ) i i d h. With d / h 0.400, we get i 357 and i 49. Thus, the current through the person is i 357, or approximately 3.6 k. 8. Line has slope 6.0 kω. Line has slope 4.0 kω. Line 3 has slope 3.0 kω. The parallel pair equivalence is /( + ).4 kω. That in series with 3 gives an equivalence of 3 + 3.4 kω+.0 kω 4.4 k Ω. The current through the battery is therefore i ε / 3 (6 )/(4.4 kω) and the voltage drop across 3 is (6 )( kω)/(4.4 kω).73. Subtracting this (because of the loop rule) from the battery voltage leaves us with the voltage across. Then Ohm s law gives the current through : (6.73 )/(4 kω) 0.8 m. 9. (a) The parallel set of three identical 8 Ω resistors reduce to 6.0 Ω, which is now in series with the 6.0 Ω resistor at the top right, so that the total resistive load across the battery is ' + Ω. Thus, the current through ' is ()/'.0, which is the current through. By symmetry, we see one-third of that passes through any one of those 8 Ω resistors; therefore, i 0.333. (b) The direction of i is clearly rightward. (c) We use Eq. 6-7: P i ' (.0 ) ( Ω) W. Thus, in 60 s, the energy dissipated is ( J/s)(60 s) 70 J. 30. Using the junction rule (i 3 i + i ) we write two loop rule equations: (a) Solving, we find i 0, and 0.0 i (i + i ) 3 0 5.00 i (i + i ) 3 0. (b) i 3 i + i.5 (downward, as was assumed in writing the equations as we did). 3. (a) We reduce the parallel pair of identical.0 Ω resistors (on the right side) to '.0 Ω, and we reduce the series pair of identical.0 Ω resistors (on the upper left side) to '' 4.0 Ω. With denoting the.0 Ω resistor at the bottom (between and ), we now have three resistors in series, which are equivalent to + + 7.0 Ω

068 CHPTE 7 across which the voltage is 7.0 (by the loop rule, this is 5.0 ), implying that the current is.0 (clockwise). Thus, the voltage across ' is (.0 )(.0 Ω).0, which means that (examining the right side of the circuit) the voltage difference between ground and is. Noting the orientation of the battery, we conclude. (b) The voltage across '' is (.0 )(4.0 Ω) 4.0, which means that (examining the left side of the circuit) the voltage difference between ground and is 5.0 + 4.0 9.0. Noting the orientation of the battery, we conclude 9.0. This can be verified by considering the voltage across and the value we obtained for. 3. (a) For typing convenience, we denote the emf of battery as and the emf of battery as. The loop rule (examining the left-hand loop) gives + i 0. Since is held constant while and i vary, we see that this expression (for large enough ) will result in a negative value for i, so the downward sloping line (the line that is dashed in Fig. 7-43(b)) must represent i. It appears to be zero when 6. With i 0, our loop rule gives, which implies that 6.0. (b) t (in the graph) it appears that i 0.. Now our loop rule equation (with the conclusion about found in part (a)) gives 0 Ω. (c) Looking at the point where the upward-sloping i line crosses the axis (at 4 ), we note that i 0. there and that the loop rule around the right-hand loop should give i i when i 0. and i 0. This leads directly to 40 Ω. 33. First, we note in 4, that the voltage across 4 is equal to the sum of the voltages across 5 and 6 : 4 i 6 ( 5 + 6 ) (.40 )(8.00 Ω + 4.00 Ω) 6.8. The current through 4 is then equal to i 4 4 / 4 6.8 /(6.0 Ω).05. By the junction rule, the current in is i i 4 + i 6.05 +.40.45, so its voltage is (.00 Ω)(.45 ) 4.90. The loop rule tells us the voltage across 3 is 3 + 4.7 (implying that the current through it is i 3 3 /(.00 Ω) 0.85 ). The junction rule now gives the current in as i i + i 3.45 + 0.85 3.3, implying that the voltage across it is (3.3 )(.00 Ω) 6.6. Therefore, by the loop rule,

069 ε + 3 6.6 +.7 48.3. 34. (a) By the loop rule, it remains the same. This question is aimed at student conceptualization of voltage; many students apparently confuse the concepts of voltage and current and speak of voltage going through a resistor which would be difficult to rectify with the conclusion of this problem. (b) The loop rule still applies, of course, but (by the junction rule and Ohm s law) the voltages across and 3 (which were the same when the switch was open) are no longer equal. More current is now being supplied by the battery, which means more current is in 3, implying its voltage drop has increased (in magnitude). Thus, by the loop rule (since the battery voltage has not changed) the voltage across has decreased a corresponding amount. When the switch was open, the voltage across was 6.0 (easily seen from symmetry considerations). With the switch closed, and are equivalent (by Eq. 7-4) to 3.0 Ω, which means the total load on the battery is 9.0 Ω. The current therefore is.33, which implies that the voltage drop across 3 is 8.0. The loop rule then tells us that the voltage drop across is 8.0 4.0. This is a decrease of.0 volts from the value it had when the switch was open. 35. (a) The symmetry of the problem allows us to use i as the current in both of the resistors and i for the resistors. We see from the junction rule that i 3 i i. There are only two independent loop rule equations: ε i i 0 ε i i i 0 ( ) 3 where in the latter equation, a zigzag path through the bridge has been taken. Solving, we find i 0.0065, i 0.005 and i 3 i i 0.000375. Therefore, B i 5.5. (b) It follows also that B C i 3 3.50. (c) We find C D i 5.5. (d) Finally, C i 6.75. 36. (a) Using the junction rule (i i + i 3 ) we write two loop rule equations: b g b g. ε i i + i3 0 ε i i + i 0 3 3 3 Solving, we find i 0.009 (rightward, as was assumed in writing the equations as we did), i 3 0.073 (leftward), and i i + i 3 0.038 (downward). (b) The direction is downward. See the results in part (a).

074 CHPTE 7 Clearly the only physically interesting solution to this is n 8. Thus, there are eight resistors in parallel (as well as that resistor in series shown toward the bottom) in Fig. 7-5. 43. Let the resistors be divided into groups of n resistors each, with all the resistors in the same group connected in series. Suppose there are m such groups that are connected in parallel with each other. Let be the resistance of any one of the resistors. Then the equivalent resistance of any group is n, and eq, the equivalent resistance of the whole array, satisfies m m. n n eq Since the problem requires eq 0 Ω, we must select n m. Next we make use of Eq. 7-6. We note that the current is the same in every resistor and there are n m n resistors, so the maximum total power that can be dissipated is P total n P, where P.0 W is the maximum power that can be dissipated by any one of the resistors. The problem demands P total 5.0P, so n must be at least as large as 5.0. Since n must be an integer, the smallest it can be is 3. The least number of resistors is n 9. 44. (a) esistors, 3, and 4 are in parallel. By finding a common denominator and simplifying, the equation / / + / 3 + / 4 gives an equivalent resistance of (50.0 )(50.0 )(75.0 ) + + (50.0 Ω)(50.0 Ω ) + (50.0 Ω)(75.0 Ω ) + (50.0 Ω)(75.0 Ω) 8.8 Ω. 3 4 Ω Ω Ω 3 4 3 4 Thus, considering the series contribution of resistor, the equivalent resistance for the network is eq + 00 Ω + 8.8 Ω 8.8 Ω 9 Ω. (b) i ε/ eq 6.0 /(8.8 Ω) 5.05 0. (c) i (ε )/ (ε i )/ [6.0 (5.05 0 )(00Ω)]/50 Ω.90 0. (d) i 3 (ε )/ 3 i / 3 (.90 0 )(50.0 Ω/50.0 Ω).90 0. (e) i 4 i i i 3 5.05 0 (.90 0 ).5 0. 45. (a) We note that the resistors occur in series pairs, contributing net resistance in each branch where they appear. Since ε ε 3 and, from symmetry we know that the currents through ε and ε 3 are the same: i i 3 i. Therefore, the current through ε is i i. Then from b a ε i ε + ( )(i) we get i ε ε 4.0.0 4+ 4(.0Ω ) +.0Ω 0.33.

075 Therefore, the current through ε is i i 0.67. (b) The direction of i is downward. (c) The current through ε is i 0.33. (d) The direction of i is upward. (e) From part (a), we have i 3 i 0.33. (f) The direction of i 3 is also upward. (g) a b i + ε (0.333 )(.0 Ω) + 4.0 3.3. 46. (a) When 3 0 all the current passes through and 3 and avoids altogether. Since that value of the current (through the battery) is 0.006 (see Fig. 7-55(b)) for 3 0 then (using Ohm s law) ( )/(0.006 ).0 0 3 Ω. (b) When 3 all the current passes through and and avoids 3 altogether. Since that value of the current (through the battery) is 0.00 (stated in problem) for 3 then (using Ohm s law) ( )/(0.00 ) 4.0 0 3 Ω. 47. (a) The copper wire and the aluminum sheath are connected in parallel, so the potential difference is the same for them. Since the potential difference is the product of the current and the resistance, i C C i, where i C is the current in the copper, i is the current in the aluminum, C is the resistance of the copper, and is the resistance of the aluminum. The resistance of either component is given by ρl/, where ρ is the resistivity, L is the length, and is the cross-sectional area. The resistance of the copper wire is C ρ C L/πa, and the resistance of the aluminum sheath is ρ L/π(b a ). We substitute these expressions into i C C i, and cancel the common factors L and π to obtain icρc iρ. a b a We solve this equation simultaneously with i i C + i, where i is the total current. We find rcρ Ci ic cr rchρc + rcρ and

079 53. The current in is i. Let i be the current in and take it to be downward. ccording to the junction rule the current in the voltmeter is i i and it is downward. We apply the loop rule to the left-hand loop to obtain ε i i ir 0. We apply the loop rule to the right-hand loop to obtain b g. i i i 0 The second equation yields i + i. We substitute this into the first equation to obtain This has the solution b + rgb+ g ε i+ i 0. i ε + r + + b gb g. The reading on the voltmeter is 3 ( 3.0) ( 5.0 0 ) ( 50 ) ( 300 00 )( 50 5.0 0 ) ( 50 )( 5.0 0 ) ε Ω Ω i + + + Ω + Ω Ω+ Ω + Ω Ω 3 3 ( r) ( ).. The current in the absence of the voltmeter can be obtained by taking the limit as becomes infinitely large. Then ε ( 3.0 )( 50Ω) + + r 50 Ω+ 300 Ω+ 00 Ω i The fractional error is (..5)/(.5) 0.030, or 3.0%. 54. (a) ε + ir + (0.0 ) (0.0500 Ω).5. (b) Now ε ' + (i motor + 8.00 )r, where.5. Therefore, ' i' light (8.00 ) (.0 /0 ) 9.60.

080 CHPTE 7 i motor ε.5 9.60 8.00 8.00 50.0. r 0.0500Ω 55. Let i be the current in and, and take it to be positive if it is toward point a in. Let i be the current in s and x, and take it to be positive if it is toward b in s. The loop rule yields ( + )i ( x + s )i 0. Since points a and b are at the same potential, i i s. The second equation gives i i / s, which is substituted into the first equation to obtain s ( + ) i ( x + s) i x. 56. The currents in and are i and i' i, respectively. Since i (i' i) we have, by dividing both sides by, (i' / i/) (/' /). Thus, s +. The equivalent resistance of the circuit is eq + 0 + + 0 +. + (a) The ammeter reading is ε ε.0 i + + + Ω+ Ω+ Ω Ω Ω+ Ω eq 0 7.09 0. (b) The voltmeter reading is ( ) 3.00 00 ( 300 ) ( 85.0 ) ( 300 85.0 ) ε i' ( + 0 ).0 (0.0709 ) (03.00 Ω) 4.70. (c) The apparent resistance is ' /i' 4.70 /(7.09 0 ) 66.3 Ω. (d) If is increased, the difference between and decreases. In fact, as. 57. Here we denote the battery emf as. Then the requirement stated in the problem that the resistor voltage be equal to the capacitor voltage becomes i cap, or e t /C ( e t/c ) where Eqs. 7-34 and 7-35 have been used. This leads to t C ln, or t 0.08 ms. 58. (a) τ C (.40 0 6 Ω)(.80 0 6 F).5 s.

08 (b) q o εc (.0 )(.80 μ F).6 μc. (c) The time t satisfies q q 0 ( e t/c ), or q.6 μc t C ( ) 0 ln.5s ln 3.40s. q0 q.6 μc 6.0 μc 59. During charging, the charge on the positive plate of the capacitor is given by q Cε e t τ c h, where C is the capacitance, ε is applied emf, and τ C is the capacitive time constant. The equilibrium charge is q eq Cε. We require q 0.99q eq 0.99Cε, so 099. e t τ. Thus, e t τ 00.. Taking the natural logarithm of both sides, we obtain t/τ ln 0.0 4.6 or t 4.6τ. 60. (a) We use q q 0 e t/τ, or t τ ln (q 0 /q), where τ C is the capacitive time constant. Thus, q 0 3 t/3 t/3 τ ln τ ln 0.4τ 0.4. q0 /3 τ ln q ln3. t.. 0 /3 (b) t/3 τ τ τ q0 /3 τ 6. (a) The voltage difference across the capacitor is (t) ε( e t/c ). t t.30 μs we have (t) 5.00, so 5.00 (.0 )( e.30 μs/c ), which gives τ (.30 μ s)/ln(/7).4 μs. (b) The capacitance is C τ/ (.4 μs)/(5.0 kω) 6 pf. 6. The time it takes for the voltage difference across the capacitor to reach L is given by ε e tc L c h. We solve for : t Cln ε ε 0500. s 6 b L g c050. 0 Fhln 950. b950. 7. 0g 6 35. 0 Ω where we used t 0.500 s given (implicitly) in the problem.

084 CHPTE 7 64. (a) The potential difference across the plates of a capacitor is related to the charge q on the positive plate by q/c, where C is capacitance. Since the charge on a discharging capacitor is given by q q 0 e t/τ, this means 0 e t/τ where 0 is the initial potential difference. We solve for the time constant τ by dividing by 0 and taking the natural logarithm: t 0. 0s τ 7. s. ln ln 00. 00 b 0g b gb g (b) t t 7.0 s, t/τ (7.0 s)/(.7 s) 7.83, so t τ 783 e 00 e 396. 0. 0 b g. 65. In the steady state situation, the capacitor voltage will equal the voltage across 5 kω: ε 0.0 0 ( 5.0 kω ).0. + 0.0 kω+ 5.0 kω Now, multiplying Eq. 7-39 by the capacitance leads to 0 e t/c describing the voltage across the capacitor (and across 5.0 kω) after the switch is opened (at t 0). Thus, with t 0.00400 s, we obtain 0. 004 b5000 0 4 0 b g ge. 6 j e 66.. Therefore, using Ohm s law, the current through is 6.6/5000 4. 0 4. 66. We apply Eq. 7-39 to each capacitor, demand their initial charges are in a ratio of 3: as described in the problem, and solve for the time. With τ C Ω 6 4 (0.0 )(5.00 0 F).00 0 s τ C Ω 6 5 (0.0 )(8.00 0 F) 8.00 0 s,

087 c hb g 7 6 iε 955. 0 400. 38. 0 W. The energy delivered by the battery is either stored in the capacitor or dissipated in the resistor. Conservation of energy requires that iε (q/c) (dq/dt) + i. Except for some round-off error the numerical results support the conservation principle. 70. (a) From symmetry we see that the current through the top set of batteries (i) is the same as the current through the second set. This implies that the current through the 4.0 Ω resistor at the bottom is i i. Thus, with r denoting the internal resistance of each battery (equal to 4.0 Ω) and ε denoting the 0 emf, we consider one loop equation (the outer loop), proceeding counterclockwise: b g b g. 3 ε ir i 0 This yields i 3.0. Consequently, i 6.0. (b) The terminal voltage of each battery is ε ir 8.0. (c) Using Eq. 7-7, we obtain P iε (3)(0) 60 W. (d) Using Eq. 6-7, we have P i r 36 W. 7. (a) If S is closed, and S and S 3 are open, then i a ε/ 0 /40.0 Ω 3.00. (b) If S 3 is open while S and S remain closed, then eq + ( + ) /( + ) 0.0 Ω + (0.0 Ω) (30.0 Ω)/(50.0 Ω) 3.0 Ω, so i a ε/ eq 0 /3.0 Ω 3.75. (c) If all three switches S, S, and S 3 are closed, then eq + '/( + ') where that is, ' + ( + )/( + ).0 Ω, eq 0.0 Ω + (0.0 Ω) (.0 Ω)/(0.0 Ω +.0 Ω) 30.5 Ω, so i a ε/ eq 0 /30.5 Ω 3.94. 7. (a) The four resistors,, 3, and 4 on the left reduce to 3 4 eq + 34 + 7.0 Ω+ 3.0 Ω 0 Ω + 3+ 4.

088 CHPTE 7 With ε 30 across eq the current there is i 3.0. (b) The three resistors on the right reduce to (6.0 Ω)(.0 Ω) + + +.5 Ω 3.0 Ω. + 6.0 Ω+.0 Ω 5 6 eq 56 7 7 5 6 With ε 30 across eq the current there is i 4 0. (c) By the junction rule, i i + i 4 3. (d) By symmetry, i 3 i.5. (e) By the loop rule (proceeding clockwise), readily yields i 5 7.5. 30 i 4 (.5 Ω) i 5 (.0 Ω) 0 73. (a) The magnitude of the current density vector is J 7.3 0 m. ( ) i 4 4 60.0 ( + ) ( + ) π D π Ω+ Ω ( 0.7 0.79 )(.60 0 3 m) (b) /( + ) (60.0 )(0.7 Ω)/(0.7 Ω + 0.79 Ω) 8.90. (c) The resistivity of wire is 3 π D π (0.7 Ω )(.60 0 m) ρ Ω L 4L 4(40.0 m) 8.69 0 m. So wire is made of copper. (d) J B 7.3 0 m. J (e) B 60.0 8.9 5.. (f) The resistivity of wire B is ρ B 968. 0 8 Ω m, so wire B is made of iron. 74. The resistor by the letter i is above three other resistors; together, these four resistors are equivalent to a resistor 0 Ω (with current i). s if we were presented with a

089 maze, we find a path through that passes through any number of batteries (0, it turns out) but no other resistors, which as in any good maze winds all over the place. Some of the ten batteries are opposing each other (particularly the ones along the outside), so that their net emf is only ε 40. (a) The current through is then i ε/ 4.0. (b) The direction is upward in the figure. 75. (a) In the process described in the problem, no charge is gained or lost. Thus, q constant. Hence, C 50 3 q C C ( 00) 3.0 0. C 0 (b) Equation 7-39, with τ C, describes not only the discharging of q but also of. Thus, 9 3000 ( ) ( ) t τ 0 0 ln 300 0 Ω 0 0 F ln 00 e t C which yields t 0 s. This is a longer time than most people are inclined to wait before going on to their next task (such as handling the sensitive electronic equipment). (c) We solve e tc 0 for with the new values 0 400 and t 0.30 s. Thus, t Cln 030. s 0 0 F ln 400 00 b 0 g c h b g 0. 0 Ω. 76. (a) We reduce the parallel pair of resistors (at the bottom of the figure) to a single.00 Ω resistor and then reduce it with its series partner (at the lower left of the figure) to obtain an equivalence of.00 Ω +.00Ω 3.00 Ω. It is clear that the current through is the i we are solving for. Now, we employ the loop rule, choose a path that includes and all the batteries (proceeding clockwise). Thus, assuming i goes leftward through, we have which yields i 5.00. 5.00 + 0.0 0.0 i 0 (b) Since i is positive, our assumption regarding its direction (leftward) was correct. (c) Since the current through the ε 0.0 battery is forward, battery is supplying energy.

09 78. The current in the ammeter is given by i ε/(r + + + ). The current in and without the ammeter is i ε/(r + + ). The percent error is then Δi i i r+ + 0.0Ω i i r+ + + r+ + +.0Ω+ 5.0Ω+ 4.0Ω+ 0.0Ω 0.90%. 79. (a) The charge q on the capacitor as a function of time is q(t) (εc)( e t/c ), so the charging current is i(t) dq/dt (ε/)e t/c. The energy supplied by the emf is then U z tc ε ε z idt e dt Cε U 0 0 C where UC Cε is the energy stored in the capacitor. (b) By directly integrating i we obtain U z ε z tc i dt e dt Cε. 0 0 80. In the steady state situation, there is no current going to the capacitors, so the resistors all have the same current. By the loop rule, 0.0 (5.00 Ω)i + (0.0 Ω)i + (5.0 Ω)i which yields i 3. Consequently, the voltage across the 5.00 Ω resistor is (5.00 Ω)(/3 ) 0/3, and is equal to the voltage across the C 5.00 μf capacitor. Using Eq. 6-, we find the stored energy on that capacitor: U 6 0 5 C (5.00 0 F).78 0 J. 3 Similarly, the voltage across the 0.0 Ω resistor is (0.0 Ω)(/3 ) 0/3 and is equal to the voltage across the C 0.0 μf capacitor. Hence, U 0 6 5 C (0.0 0 F). 0 J 3

09 CHPTE 7 Therefore, the total capacitor energy is U + U.50 0 4 J. 8. The potential difference across is i ε + + 3 b gb4. 0Ωg 40.. 30. Ω+ 40. Ω+ 50. Ω 8. From a ε c ir i and i (ε ε )/( + r + r ), we get ε ε a c ε i( r+ ) ε ( r+ ) + r+ r 4.4. 4.4 (.3Ω+ 5.5 Ω) 5.5Ω+.8Ω+.3Ω.5. 83. The potential difference across the capacitor varies as a function of time t as t/ C t () e 0. Thus, t. C ln ( ) 0 (a) Then, for t min 0.0 μs, min 0.0 μs 4.8 Ω. ( 0.0 μf) ln ( 5.00 0.800) (b) For t max 6.00 ms, F H G I K J b g 600. ms max 0. 0 μs 4. 8Ω 49. 0 4 Ω, where in the last equation we used τ C. 84. (a) Since i ( ) tank 40 Ω, 0Ω+ 40Ω 8.0 0. (b) Now, tank (40 Ω + 0 Ω)/ 80 Ω, so i /(0 Ω + 80 Ω) 0.3. (c) When full, tank 0 Ω so i /(0 Ω + 0 Ω) 0.40. 85. The internal resistance of the battery is r (.4 )/50 0.0 Ω < 0.00 Ω, so the battery is OK. The resistance of the cable is so the cable is defective. 3.0 /50 0.060 Ω > 0.040 Ω,