Chapter 8 - Direct Current Circuits 8. (a = becomes Δ V = 0.0 W = so = 6.7 Ω (.6 V so.6 V = ( 6.7 Ω FG. 8. and =.7 A = + r so 5.0 V =.6 V + (.7 A r r =.97 Ω.00 V 8. The total resistance is = = 5.00 Ω. 0.600 A (a lamp = r batteries = 5.00 Ω 0.408 Ω = 4.59 Ω batteries total ( 0.408 Ω ( 5.00 Ω 8. (a Here ( = = 0.08 6 = 8.6%.6 V = + r, so = = =.48 A. + r 5.00 Ω+ 0.080 0 Ω Then, Δ V = =.48 A 5.00 Ω =.4 V. et and be the currents flowing through the battery and the headlights, respectively. Then, = + 5.0 A, and r r = 0 = + 5.0 A 0.080 0 Ω + 5.00 Ω =.6 V so giving =.9 A Thus, V Δ =.9 A 5.00 Ω = 9.65 V FG. 8. FG. 8.
8.7 f we turn the given diagram on its side, we find that it is the same as figure (a. The 0.0 Ω and 5.00 Ω resistors are in series, so the first reduction is shown in. n addition, since the 0.0 Ω, 5.00 Ω, and 5.0 Ω resistors are then in parallel, we can solve for their uivalent resistance as: = =.94 Ω ( + + 0.0 Ω 5.00 Ω 5.0 Ω This is shown in figure (c, which in turn reduces to the circuit shown in figure (d. Δ Next, we work backwards through the diagrams applying = V and Δ V = alternately to every resistor, real and uivalent. The.94 Ω resistor is connected across 5.0 V, so the current through the battery in every diagram is ΔV 5.0 V = = =.9 A.94 Ω n figure (c, this.9 A goes through the.94 Ω uivalent resistor to give a potential difference of: Δ V = =.9 A.94 Ω = 5.68 V From figure, we see that this potential difference is the same across V, the 0 Ω resistor, and the 5.00 Ω resistor. Δ ab Thus we have first found the answer to part, which is Δ V ab = 5.68 V. (a Since the current through the 0.0 Ω resistor is also the current through the 5.0 Ω line ab, ΔVab 5.68 V = = = 0.7 A = 7 ma 5.0 Ω ab FG. 8.7 8.0 Using.00-Ω,.00-Ω, and 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series.00 Ω 6.00 Ω.00 Ω 7.00 Ω 4.00 Ω 9.00 Ω 5.00 Ω arallel Mixed 0.9 Ω.56 Ω.0 Ω.00 Ω. Ω. Ω.7 Ω.7 Ω 4. Ω 5.0 Ω FG. 8.0 *8. The resistance between a and b decreases. Closing the switch opens a new path with resistance of only 0 Ω. The original resistance is + = + 50 Ω. + 90 + 0 0 + 90
+ + = + 8 + + 90 0 0 90 The final resistance is Ω. We ruire + 50 Ω = ( + 8 Ω so = 4.0 Ω 8.5 s battery p = + = 0.750 Ω.00.00 =.00 + 0.750 + 4.00 Ω = 6.75 Ω = ΔV 8.0 V = = =.67 A 6.75 Ω s : ( =.67 A.00 Ω = 4. W in.00 Ω =.67 A 4.00 A = 8.4 W in 4.00 Ω Δ V =.67 A.00 Ω = 5. V, Δ V =.67 A 4.00 Ω = 0.67 V (.00 V (.00 V ( V Δ V = 8.0 V ΔV Δ V =.00 V =Δ V =Δ 4 4 p 4 = = =. W in.00 Ω.00 Ω = = = 4.00 W in.00 Ω.00 Ω FG. 8.5 8.6 + 5.0 7.00.00 5.00 = 0 5.00 = 7.00 so = 0.74 A = + =.00 A 0.74 + =.00 so =.9 A FG. 8.6 +.00.9 5.00.00 = 0 =.6 V * 8.0 (a The first uation represents Kirchhoff s loop theorem. We choose to think of it as describing a clockwise trip around the left-hand loop in a circuit; see Figure (a. For the right-hand loop see Figure. The junctions must be between the 5.8 V and the 70 Ω and between the 70 Ω and the 50 Ω. Then we have Figure (c. This is consistent with the third uation, 0 Ω 70 Ω 5.8 V 70 Ω Figure (a 50 Ω Figure.0 V
We substitute: + = 0 = + 0 + 5.8 70 70 = 0 + 70 + 70 + 50. = 0 5.8 V 50 Ω 0 Ω 70 Ω.0 V Figure (c Next FG. 8.0 5.8 590 = 70 50 70 + ( 5.8 590. = 0 70 70 + 8.5 89. = 0 5.05 V = = 459 Ω.0 ma in the 0- Ω resistor and out of the positive pole of the 5.8-V battery 5.8 590 0.0 0 = =.87 ma 70 The current is.87 ma in the 50- Ω resistor and out of the negative pole of the.-v battery. =.0.87 = 9. ma in the 70- Ω resistor *8. et 6 represent the current in the ammeter and the top 6-Ω resistor. The bottom 6-Ω resistor has the same potential difference across it, so it carries an ual current. For the top loop we have 6 V 0 Ω 0 6 Ω = 0 6 For the bottom loop, 4.5 5 6 = 0. 5 6 For the junctions on the left side, taken together, + 0 + 5 = 0. 6 6 We eliminate 0 = 0.6 0.6 and = 0.9. by substitution: 6 5 6 0.6 0.6 + 0.9. = 0 =.5/.8 = 0.95 A 6 6 6 6 The loop theorem for the little loop containing the voltmeter gives + 6 V ΔV 4.5 V = 0 ΔV =.50 V 8. abel the currents in the branches as shown in the first figure. educe the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff s loop rule to both loops in Figure to obtain: and.7 +.7 = 50.7 +.7 = 500 (a With =000 Ω, simultaneous solution of these uations yields: = 0.0 ma
and = 0.0 ma From Figure, Thus, from Figure (a, Vc Va = +.7 = 40 V 4 Vc Va 40 V = = = 60.0 ma 4 4000 Ω Finally, applying Kirchhoff s point rule at point a in Figure (a gives:, = 4 = 60.0 ma 0.0 ma = + 50.0 ma FG. 8. or = 50.0 ma from point ato point e 8.4 Using Kirchhoff s rules,.0 ( 0.00 0 ( 0.060 0 = 0 0.0 + (.00 ( 0.060 0 = 0 and = +.0 ( 0.00 0 ( 0.070 0 = 0 0.0 +.00 0.060 0 = 0 Solving simultaneously, = 0.8 A dow nw ard in the dead battery and = 7 A downward in the starter FG. 8.4 The currents are forward in the live battery and in the starter, relative to normal starting operation. The current is backward in the dead battery, tending to charge it up. 8. (a Call the potential at the left junction V and at the right V. After a long time, the capacitor is fully charged. V = 8.00 V because of voltage divider: 0.0 V = =.00 A 5.00 Ω V = 0.0 V.00 A.00 Ω = 8.00 V FG. 8.(a ikewise, or Therefore, V.00 Ω = ( 0.0 V =.00 V.00 Ω+ 8.00 Ω 0.0 V = =.00 A 0.0 Ω V = ( 0.0 V ( 8.00 Ω (.00 A =.00 V Δ V = V V = 8.00.00 = 6.00 V
edraw the circuit and = =.60 Ω ( 9.00 Ω + ( 6.00 Ω C = e tc 6.60 0 s = 0 so t = Cln 0 = 8.9 μs FG. 8. 8. (a (c 5 6 τ = C =.50 0 Ω 0.0 0 F =.50 s 5 6 τ =.00 0 Ω 0.0 0 F =.00 s The battery carries current The 00 kω carries current of magnitude So the switch carries downward current 0.0 V = 00 μa 50.0 0 Ω 0.0 V e e 00 0 Ω tc t.00 s = 0 = 00 μa + ( 00 μa e t.00 s 8.5 Series esistor Voltmeter Δ V = : 5.0 =.50 0 + 75.0 Solving, s = 6.6 kω s FG. 8.5 *8.40 5 V The current in the battery is =.5 A. 0 Ω+ + 5 Ω 8 Ω The voltage across 5 Ω is 5 V 0 Ω.5 A =.5 V. (a The current in it is.5 V/5 Ω = 0.706 A. =.5 V 0.706 A =.49 W (c Only the circuit in Figure 8.40c ruires the use of Kirchhoff's rules for solution. n the other circuits the 5-Ω and 8-Ω resistors are still in parallel with each other. (d The power is lowest in Figure 8.40c. The circuits in Figures 8.40b and 8.40d have in effect 0-V batteries driving the current. ( + r *8.4 = = = + r, so or et + r = x, then ( + r = x or + r x r = 0 With r =.0 Ω, this becomes +.40 x.44 = 0
which has solutions of ( x ( x.40 ±.40 5.76 = (a With = 9.0 V and =.8 W, x = 6.6 : + 4.± 4. 5.76 = =.84 Ω or 0.75 Ω. Either external resistance extracts the same power from the battery. For = 9.0 V and =. W, x =.99 +.59 ±.59 5.76.59 ±. = = The uation for the load resistance yields a complex number, so there is no resistance that will extract. W from this battery. The maximum power output occurs when = r =.0 Ω, and that maximum is max = = 7.6 W. 4r *8.45 The charging current is given by 4.7 V. V 0.85 Ω = 0 =.76 A The energy delivered by the 4.7 V supply is ΔVt = 4.7 V.76 A.80 h ( 600 s/h = 68 000 J J The energy stored in the battery is. V.76 A.8 ( 600 s = 5 000 J The same energy is released by the emf of the battery:. V ( 7. ( 600 s = 5 000 so the discharge current is = 0.45 A The load resistor is given by. V (0.45 A (0.45 A 0.85 Ω = 0 = (.8 V/0.45 A = 9.5 Ω The energy delivered to the load is ΔVt = t = (0.45 A 9.5 Ω (7. 600 s = 47 000 J The efficiency is 47 000 J/68 000 J = 0.87 *8.50 (a When the capacitor is fully charged, no current exists in its branch. The current in the left resistors is 5 V/8 Ω = 0.060 A. The current in the right resistors is 5 V/( Ω +. elative to the negative side of the battery, the left capacitor plate is at voltage 80 Ω (0.060 A = 4.8 V. The right plate is at (5 V/( Ω +. The voltage across the capacitor is 4.8 V (5 V/( Ω +. The charge on the capacitor is
Q = μf [4.8 V (5 V/( Ω + ] = (8.9 Ω 0.54 μc/( Ω + With = 0 Ω, Q = (8.9 5.4 μc/( + 0 =.96 μc (c Yes. Q = 0 when 8.9 Ω 0.54 = 0 = 5. Ω (d The maximum charge occurs for = 0. t is 8.9/ = 4.5 μc. (e Yes. Taking = corresponds to disconnecting a wire to remove the branch containing. n this case Q = 0.54 / = 0.54 μc. 8.5 (a tc ( q= CΔV e 6 6 6 0.0.00 0.00 0 q= (.00 0 F( 0.0 V e = 9.9 μc dq ΔV = = e dt tc 0.0 V e.00 0 Ω 5.00 8 = = = 6.7 0 A.7 na (c du d q q dq q = = = dt dt C C dt C du dt.7 0 A.4 0 W 4 nw.00 0 C V 6 9.9 0 C 8 7 = ( = = 6 (d 8 7 battery = =.7 0 A 0.0 V =.7 0 W = 7 nw The battery power could also be computed as the sum of the instantaneous powers delivered to the resistor and to the capacitor: + du/dt = (.7 0 9 A 0 6 Ω + 4 nw = 7 nw *8.54 (a We find the resistance intrinsic to the vacuum cleaner: = Δ V = ( 0 V = = = 6.9 Ω 55 W with the inexpensive cord, the uivalent resistance is 0.9 Ω+ 6.9 Ω+ 0.9 Ω= 8.7 Ω so the current throughout the circuit is FG. 8.54 0 V = = = 4.8 A 8.7 Ω Tot and the cleaner power is
cleaner = Δ V = = 4.8 A 6.9 Ω = 470 W cleaner n symbols, = + r, Tot = + r and = = ( + r cleaner (c 6.9 Ω + r = = 0 V = 7. Ω cleaner 55 W 7. Ω 6.9 Ω ρl ρl4 r = = 0.8 Ω = = A π d 8 4ρl 4.7 ( 0 Ω m( 5 m d = = =.60 mm or more π r π ( Ω 0.8 Unless the extension cord is a superconductor, it is impossible to attain cleaner power 55 W. To move from 55 W to 5 W will ruire a lot more copper, as we show here: 0 V 6.9 Ω 6.9 Ω r = = = 0.07 9 Ω cleaner 5 W 8 4ρl 4.7 ( 0 Ω m( 5 m d = = =.9 mm or more π r π ( Ω 0.07 9 8.55 (a First determine the resistance of each light bulb: = ( 0 V = = = 40 Ω 60.0 W We obtain the uivalent resistance of the network of light bulbs by identifying series and parallel uivalent resistances: FG. 8.55 = + = 40 Ω+ 0 Ω = 60 Ω + The total power dissipated in the 60 Ω is ( 0 V = = = 60 Ω 40.0 W The current through the network is given by = : 40.0 W = = = A 60 Ω The potential difference across Δ V = = A ( 40 Ω = 80.0 V is The potential difference Δ across the parallel combination of and is V
Δ V = = A = ( Ω + ( Ω 40 40 40.0 V 8.59 From the hint, the uivalent resistance of. That is, T + = + T + = + + + = + T T = 0 Only the + sign is physical: For example, if T = Ω T T 4 ( ( = T ± T T ( 4 T T T = + + And = 0 Ω, = 5 Ω 8.6 (a After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for : = 0 steady-state For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the -kω and 5-kΩ resistors in series: For and : 9.00 V + = = = + (.0 kω+ 5.0 kω ( μa steady-state After the transient currents have ceased, the potential difference across C is the same as the potential = because there is no difference across voltage drop across. Therefore, the charge Q on C is ( μ ( μ Q = C Δ V = C = 0.0 F A 5.0 kω = 50.0 μc FG. 8.6
(c When the switch is opened, the branch containing is no longer part of the circuit. The capacitor discharges through + with a time constant of ( μ + C = 5.0 kω+.00 kω 0.0 F = 0.80 s. The initial current in this discharge circuit is i determined by the initial potential difference across the capacitor applied to + in series: FG. 8.6(c i ( ( μa ( 5.0 kω C ( + ( + ( 5.0 kω+.00 kω = = = = 78 μa Thus, when the switch is opened, the current through changes instantaneously from μa (downward to 78 μa (downward as shown in the graph. Thereafter, it decays according to ( μ ( 0.80 s = e = 78 A e for t > 0 t ( + C t i Q (d The charge q on the capacitor decays from Q to i i according to 5 t ( + C q= Qe i Qi ( t 0.80 s = Qe i 5 t 0.80 s 5 = e t ln 5= 80 ms t = 0.80 s ln 5 = 90 ms