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Unformatted text preview: 25.32. IDENTIFY: When current passes through a battery in the direction from the — terminal toward the + terminal, the
terminal voltage Vab of the battery is Vab = 8 — Ir . Also, Vab : IR, the potential across the circuit resistor. SET UP: 8 = 24.0 V. I =4.00 A. E—Vab _ 24.0 V~21.2 v EXECUTE: (a) Va =8alr gives r = =0 7 Q
” 1 4.00 A ' 00 '
(b) Vab ~IR=0 so szi’: 21'2 V =5.30 Q.
I 4.00 A
EVALUATE: The voltage drop across the internal resistance of the battery causes the terminal voltage of the 24.0 V battery to be less than its emf. The total resistance in the circuit is R + r = 6.00 9 . I = = 4.00 A which
6.00 ’ agrees with the value speciﬁed in the problem. 25.35. IDENTIFY: The terminal voltage of the battery is Vab = 8 — If. The voltmeter reads the potential difference between its temiinals.
SET UP: An ideal voltmeter has infinite resistance.
EXECUTE: (a) Since an ideal voltmeter has inﬁnite resistance, so there would be NO current through the 2.0 Q resistor.
(b) Vab = 8 = 5.0 V; since there is no current there is no voltage lost over the internal resistance. (0) The voltmeter reading is therefore 5.0 V since with no current ﬂowing there is no voltage drop across either resistor.
EVALUATE: This not the proper way to connect a voltmeter. If we wish to measure the terminal voltage of the battery in a circuit that does not include the voltmeter, then connect the voltmeter across the terminals of the battery.
25.36. IDENTIFY: The sum of the potential changes around the circuit loop is zero. Potential decreases by IR when
going through a resistor in the direction of the current and increases by 8 when passing through an emf in the direction from the — to + terminal.
SET UP: The current is counterclockwise, because the 16 V battery determines the direction of current ﬂow. EXECUTE: +16.0 V—8.0 V—I(l.6 Q+5.0 Q+1.4 Q+9.0 Q)=0
I: 16.0V—8.0V
v 1.6Q+5.0Q+1.4Q+9.0Q
(b) Vb +16.0 V—I(l.6 S2) =Va , so Va —Vb =Vab = 16.0 V—(1.6 £2)(0.47 A) = 15.2 V.
(0) VC +8.0 V+I(1.4 Q+5.0 Q) =Va so Vac = (5.0 Q)(O.47 A)+(1.4 S2)(0.47 A)+8.0 V =11.0 V. (d) The graph is sketched in Figure 25.36.
EVALUATE: Vd, = (0.47 A)(9.0 Q) = 4.2 V. The potential at point b is 15.2 V below the potential at point a and the potential at point c is 11.0 V below the potential at point a, so the potential of point c is
15.2 V —11.0 V = 4.2 V above the potential of point b. 16V = 0.47 A b a c Figure 25.36 26.4. 26.5. 26.11. IDENTIFY: For resistors in parallel the voltages are the same and equal to the voltage across the equivalent
resistance.
SETUP: V=IR. i=£~+~L
Req R1 R2
1 1 “
EXECUTE: (a) Re 2 ~——— + ———— = 12.3 Q.
q 32 Q 20 £2
(b)1=_V_=E£)X=19,5 A,
Raq 12.3 9 V 240 V V 240 V
c I =——=——=7.5 A;I =——=————212A.
0 m R 329 2°“ R 209 EVALUATE: More current flows through the resistor that has the smaller R.
IDENTIFY: The equivalent resistance will vary for the different connections because the seriesparallel combinations vary, and hence the current will vary. _
SET UP: First calculate the equivalent resistance using the seriesparallel formulas, then use Ohm’s law (V = R1) to find the current.
EXECUTE: (a) UK = 1/(15.0 Q) + 1/(30.0 9) gives R = 10.0 9. I = V/R = (35.0 V)/(10.0 Q) = 3.50 A. (b) 1/R = 1/(100 o) + 1/(350 :2) gives R = 7.78 9.1 = (35.0 V)/(7.78 Q) = 4.50 A
(1:) UR = 1/(200 Q) + 1/(250 9) gives R = 11.11 (2, so I: (35.0 V)/(11.11 Q) = 3.15 A. ' ' ‘ ' he 3.0042 internal resistance of the battery
((1) From part (b), the reSIStance of the triangle alone IS 7.78 S2. Adding t ' V _
gives an equivalent resistance for the circuit of 10.78 9. Therefore the current is I = (35.0 V)/(10.78 Q) — 3.25 A EVALUATE: It makes a big difference how the triangle is connected to the battery. IDENTIFY: For resistors in parallel, the voltages are the same and the currents add. i =L+—1— so Req = R‘R2 ,
Req R1 R2 R, + R2 For resistors in series, the currents are the same and the voltages add. Req = R1 + R2 . SET UP: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits
shown in Figure 26.11. EXECUTE: Req = 5.00 £2 . In Figure 26.11c, I = 2; = 12.0 A . This is the current through each of the
resistors in Figure 26.11b. Vl2 = IRl2 = (12.0 A)(2.00 Q) = 24.0 V . V34 = IR34 = (12.0 A)(3.00 9) =36.0 V . Note
that Vl2 +V34 =60.0 V . V12 is the voltage across R1 and across R2 , so Il =£= 24'0 V £8.00 A and
R1 3.00 (2
I2 =E = 24‘0 V = 4.00 A . v34 is the voltage across R3 and across R4 , so 13' = 51 = M = 3.00 A and
R2 6.00 Q R3 12.0 $2
I4 =Yﬂ= 36'0 V =9.00 A
R4 4.00 9
EVALUATE: Note that 11 + I2 2 13 + 14.
8
g 6‘
R1 R3 7
R2 R4 R12 R34 R“!
(a) (b) (C) Figure 26.11 26.13. IDENrIFY: In both circuits, with and without R4, replace series and parallel combinations of resistors by their
equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and
voltages in the original circuit. Use P : I 2R to calculate the power dissipated in each bulb. (3) SET UP: The circuit is sketched in Figure 26.13a. EXECUTE: R2, R3, and R4 are in parallel, so their equivalent resistance 1 1 1 1
R ‘ 'venb ————=—+—+——
sq 1sg1 Y Rm R2 R3 R4 ﬁgure 26.13a ; and Req =1.5o o. Figure 26.13b _ 9.00 v
4.50 o+1.50 :2 Then Vl = I.Rl = (1.50 A)(4.50 o) = 6.75 v I... =150 A, Veq =I.,qReq =(1.50 A)(1.50 Q)=2.25 v For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so
V2 =V3 =V4 =2.25 V. 1 =1.50Aand1,=1.50A 2 =ﬁ=m=0500 A, 13 =Zl=0500 A, 14 =£=O.SOOA
1e2 4.509 R3 R4 EVALUATE: Note that I2 + I3 + 14 = 1.50 A, which is I eq. For resistors in parallel the currents add and their sum is the current through the equivalent resistor.
(b) SET UP: P = 12R EXECUTE: I; = (1.50 A)2(4.50 :2) =10.1 w
P2 = P3 = P4 = (0.500 A)2 (4.50 9) =1.125 W, which rounds to 1.12 W. Rl glows brightest. EVALUATE: Note that P2 +P3 + I} = 3.37 W. This equals Pa] = Ing = (1.50 A)2(1.50 Q) = 3.37 W, the power
dissipated in the equivalent resistor. (c) SET UP: With R4 removed the circuit becomes the circuit in Figure 26.130. EXECUTE: R2 and R3 are in parallel and
their equivalent resistance Req is given by .L:i+i= 2 andReq=2.259
R R2 R3 4.509 9‘] Figure 26.13c
The equivalent circuit is shown in Figure 26.13d.
5—1091 +qu)=0 1= ’5
n+1?“l
1=—i00_v__=1.333A
4.50r2+2.25r2 Figure 26.13d
11:1.33 A, Vl =I,Rl =(l.333 A)(4.50 £2) = 6.00 V Ieq =1.33 A, Veq =quReq =(1.333 A)(2.25 (2) =3.00 v and V2 =t/3 =3.00 V.
V2 _ 3.00 v V 12 = _. _
R2 4.50 9 (d) SET UP: P = IZR' EXECUTE: I} = (1.333 A)2(4.50 9) =8.00 W P2 = P3 = (0.667 A)2(4.50 £2) = 2.00 w. (e) EVALUATE: When R4 is removed, P1 decreases and P2 and P3 increase. Bulb R1 glows less brightly and bulbs
R2 and R3 glow more brightly. When R4 is removed the equivalent resistance of the circuit increases and the current
through R1 decreases. But in the parallel combination this current divides into two equal currents rather than three,
so the currents through R2 and R3 increase. Can also see this by noting that with R4 removed and less current
through R1 the voltage drop across R1 is less so the voltage drop across R2 and across R3 must become larger. 20.667 A, 13 =E3= 0.667 A 26.14. IDENTIFY: Apply Ohm's law to eachresistor. SET UP: For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents
are the same and the voltages add. EXECUTE: From Ohm’s law, the voltage drop across the 6.00 9 resistor is V: [R = (4.00 A)(6.00 £2)  24.0 V. calculate the emf of the battery.
26.15. IDENTIFY: Apply Ohm's law to each resistor. SET UP: For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents
are the same and the voltages add. EXECUTE: The current through 2.00—S2 resistor is 6.00 A. Current through LOOQ resistor also is 6.00 A and the
voltage is 6.00 V. Voltage across the 6.00S2 resistor is 12.0 V + 6.0 V = 18.0 V. Current through the 6.00—S2
resistor is (18.0 V)/(6.00 Q) = 3.00 A. The battery emf is 18.0 V. EVALUATE: The current through the battery is 6.00 A + 3.00 A = 9.00 A. The equivalent resistor of the resistor
network is 2.00 S2, and this equals (18.0 V)/(9.00 A). ...
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This homework help was uploaded on 04/09/2008 for the course PH 1120 taught by Professor Keil during the Spring '08 term at WPI.
 Spring '08
 Keil

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