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Unformatted text preview: i tiger 7‘}? z // ; 26.21. EDENTIFY: Apply Kirchhoffs point rule at point a to find the current through R. Apply Kirchhost loop rule to
loops (1) and (2) shown in Figure 26.2121 to calculate R and 5. Travel around each loop in the direction shown. (3) SET UP: Figure 26.21a EXECUTE: Apply Kirchhoffs point rule to point a: Z] = 0 so I + 4.00 A ~ 6.00 A = 0 I , I = 2.00 A (in the direction shown in the diagram).
(b) Apply Kirchhost loop rule to loop ( 1): —(6.00 A)(3.00 Q) — (2.00 A)R + 28.0 V = 0 —18.0 V—(2.00 o)R+28.0 V=0 . R=28'0V_18'OV=5.00§2
2.00A (c) Apply Kirchhoff's loop rule to loop (2): —(6.00 A)(3.00 Q) — (4.00 A)(6.00 Q) + 8 = 0 8 = 18.0 V+24.0 V = 42.0 V
EVALUATE: Can check that the loop rule is satisﬁed for loop (3), as a check of our work: 28.0 V ~5+(4.00 A)(6.00 £2) ~(2.00 A)R =0 28.0 V —42.0 V + 24.0 V —(2.00 A)(5.00 s2) = 0 52.0 V = 42.0 V + 10.0 V
52.0 V = 52.0 V, so the loop rule is satisﬁed for this loop. ((1) IDENTIFY: If the circuit is broken at point x there can be no current in the 6.00 S2 resistor. There is now only a single current path and we can apply the loop rule to this path.
SET UP: The circuit is sketched in Figure 26.21b. Figure 26.21!)
EXECUTE: +280 V —(3.00 (2)] —(5.00 9)] = 0
I = 28'0 V =3.50 A
8.00 9 EVALUATE: Breaking the circuit at x removes the 42.0 V emf from the circuit and the current through the
3.00 (2 resistor is reduced. 26.22. IDENTHW: Appiy the loop rule andjunction rule.
SET UP: The circuit diagram is given in Figure 26.22. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. EXECUTE: The loop rule applied to loop (1) gives:
+200 V —(1.00 A)(1.00 S2)+ (1.00 A)(4.00 Q)+(l.00 A)(1.00 £2)—5l —(1.00 A)(6.00 £2) =0 61 = 20.0 V ~ 1.00 V +4.00 V +1.00 V —6.00 V : 18.0 V. The loop rule applied to loop (2) gives: +200 V — (1.00 A)(l.00 £2)—— (2.00 A)(1.00 Q) —62 — (2.00 A)(2.00 Q) — (1.00 A)(6.00 £2) = 0 £2 = 20.0 V — 1.00 V — 2.00 V — 4.00 V — 6.00 V = 7.0 V . Going from b to a along the lower branch, Vb +(2.00 A)(2.00 Q) + 7.0 V +(2.00 A)(1.00 £2) = Va . Vb —Va 2 —13.0 V ; point I) is at 13.0 V lower potential than point a.
EVALUATE: We can also calculate Vb — V“ by going from b to a along the upper branch of the circuit. Vb —(1.00 A)(6.00 Q) + 20.0 V ~(1.00 A)(1.00 Q) 2 Va and Vb —Va 2 —l3.0 V. This agrees with Vb —Va calculated
along a different path between 12 and a. 1.00 a 20“) V 6.00 o Figure 26.22 26.23. IDENTIFY: Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop
rule to three loops to calculate 81, 82 and R. ’
(a) SET UP: The circuit is sketched in Figure 26.23. Figure 26.23 EXECUTE: Apply the junction rule to point a: 3.00 A +5.00 A — I3 = 0 I3 = 8.00 A Apply the junction rule to point I): 2.00 A + I 4 — 3.00 A = 0 I 4 = 1.00 A Apply the junction rule to point c: I3 — I 4 ~ [5 = 0 I5 = I3 —14 =8.00 A—1.00 A =7.00 A EVALUATE: As a check, apply the junction rule to point d: Is — 2.00 A — 5.00 A = 0
15 = 7.00 A 26.24. 26.38. 26.39. (h) EXEUJTE: Apply the loop rule to loop (1): S. —(3.00 A)(4.00 S2) _ 13 (3.00 £2) : 0
51:12.0 v + {8.00 A)(3.00 g2) :3e.0 v Apply the loop rule to loop (2): 52 ‘(500 A)(6.00 Q)—13 (300 S2) : 0 £2 : 30.0 v +(8.00 A)(3.00 o) = 54.0 v (c) Apply the loop rule to loop (3): —(2.00 A)R ~61 +52 = 0 Rzii:54‘ov_36'ov:9.oog 2.00 A 2.00 A
EVALUATE: Apply the loop rule to loop (4) as a check of our calculations: —(2.00 A)R~(3.00 A)(4.00 o)+(5.00 A)(6.00 £2) =0
—(2.00 A)(9.00 o)—12.0 V +300 v =0 ~18.0 V + 18.0 V = 0
IDENTIFY: Use Kirchhoff’ 3 Rules to find the currents.
SET UP: Since the 1.0 V battery has the larger voltage, assume II is to the left through the 10 V battery, I2 is to the right through the 5 V battery, and I3 is to the right through the 10 Q resistor. Go around each loop in the counterclockwise direction.
EXECUTE: Upper loop: 10.0 V —(2.00 9 + 3.00 (2)1l —— (1.00 (2 + 4.00 (2)12 —5.00 V = 0 . This gives 5.0v—(5.00s2)11—'(5.00r2)12 =0 , and :>Il+12 =1.00A.
Lower loop: 5.00 V+(1.00Q+4.00 9)]2 ~(10.0 82)]3 =0 . This gives 5.00 V+(5.00S2)12 —(10.0 Q)I3 =0 , and I2 —— 213 = —1.00 A
Along with I1 = I2 + [3, we can solve for the three currents and ﬁnd: 11 = 0.800 A,12 = 0.200 A, 13 = 0.600 A. (b) Vab = —(0.200 A)(4.00 o)—(0.800 A)(3.00 o) =—3.20 v. EVALUATE: Traveling from b to a through the 4.00 O and 3.00 9 resistors you pass through the resistors in the
direction of the current and the potential decreases; point b is at higher potential than point a. IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each time. SET UP: The voltage across a capacitor is VC = q/ C . EXECUTE: (a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge
stored. (b) Since VC = 0, the full battery voltage appears across the resistor VR = S = 125 V. (c) There is no charge on the capacitor. ((1) The current through the resistor is i = J?— = 125 V
Rtotal 7500 (2 (e) After a long time has passed the full battery voltage is across the capacitor and i: 0 . The voltage across the
capacitor balances the emf: VC =125 V. The voltage across the resister is zero. The capacitor’s charge is q = CVC = (4.60x10‘6 F) (125 V) = 5.75 x104 C. The current in the circuit is zero.
EVALUATE: The current in the circuit starts at 0.0167 A and decays to zero. The charge on the capacitor starts at zero and rises to q = 5.75x104 C.
IDENTIFY: The capacitor discharges exponentially through the voltmeter. Since the potential difference across
the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases exponentially with the same time constant as the charge.
SET UP: The reading of the voltmeter obeys. the equation V = Voe“’RC, where RC is the time constant. EXECUTE: (3) Solving for C and evaluating the result when t: 4.00 5 gives = 0.0167 A. t 4.00 s
C=———=———————— = 8.49x10'7F
R ln(V/Vo) (3.4%“). mln(12.0v
3.00 v (b) 1: RC: (3.40 x 106 Q)(8.49 x 10'7 F) = 2.89 s
EVALUATE: In most laboratory circuits, time constants are much shorter than this one. 26.67. IDENTIFY: In Figure 26.67, points a and c are at the same potential and points d and b are at the same potential,
so we can calculate Vab by calculating Ved. We know the current through the resistor that is between points 0 and d. We thus can calculate the terminal voltage of the 24.0 V battery without calculating the current through it. SET UP:
II = 0.0706 A 200 Q 7t Figure 26.67 EXECUTE: V, + 11 (10.0 9) + 12.0 V = VC Vt —Vd =12.7 V; Va —Vb =Vc —Vd =12.7 V EVALUATE: The voltage across each parallel branch must be the same. The current through the 24.0 V battery
must be (24.0 V ——12.7 V)/(10.0 Q) = 1.13 A in the direction b to a. 26.71. IDENTIFY and SET UP: For part (a) use that the full emf is across each resistor. In part (b), calculate the power
dissipated by the equivalent resistance, and in this expression express RI and R2 in terms of PI, P2 and 8. EXECUTE: P1 = 52 /R1 so R1: 82/131 P2 =82/RZ so R2 =82/P2 (a) When the resistors are connected in parallel to the emf, the voltage across each resistor is 8 and the power
dissipated by each resistor is the same‘as if only the one resistor were connected. Rm = P1 + P2 (b) When the resistors are connected in series the equivalent resistance is Req = R‘ + R2 .92 62 RP. P = =__
‘°‘ R,+R2 £2/R+£2/P2 P,+P2 F]— = i + —1— Our results are that for parallel the powers add P P
lot 1 2
and that for series the reciprocals of the power add. This is opposite the result for combining resistance. Since P = £2 / R tells us that P is proportional to l/R, this makes sense. ,
26.72. IDENTIFY and SET UP: Just after the switch is closed the charge on the capacitor is zero, the voltage across the
capacitor is zero and the capacitor can be replaced by a wire in analyzing the circuit. After a long time the current to the capacitor is zero, so the current through R3 is zero. After a long time the capacitor can be replaced by a break EVALUATE: The result in part (b) can be written as in the circuit. _
EXECUTE: (a) Ignoring the capacitor for the moment, the equivalent resistance of the two parallel resistors is 1— = ”L— + L = 3 ; R = 2.00 £2 . In the absence of the capacitor, the total current in the circuit (the
R 6.00 9 3.00 .0 6.00 (2 W . current through the 8.00 9 resistor)‘would be i = £ = ﬂ = 4.20 A , of which 2/3 , or 2.80 A, would
R 8.00 (2 + 2.00 9 go through the 3.00 9 resistor and 1/ 3 , or 1.40 A, would go through the 6.00 9 resistor. Since the current . . . . V _ . . . .
through the capacrtor rs grven by r = Xe ”RC, at the instant t = O the crrcurt behaves as through the capacrtor were not present, so the currents through the various resistors are as calculated above.
(b) Once the capacitor is fully charged, no current ﬂows through that part of the circuit. The 8.00 9 and the 6.00 S2 resistors are now in series, and the current through them is i = 8/ R = (42.0 V) /(8.00 Q+6.00 Q) = 3.00 A.
The voltage drop across both the 6.00 9" resistor and the capacitor is thus V = iR = (3.00 A)(6.00 Q) = 18.0 V.
(There is no current through the 3.00 (2 resistor and so no voltage drop across it.) The charge on the capacitor is Q = CV = (4.00x10~6 F)(18.0 V): 7.2x10‘5 C.
EVALUATE: The equivalent resistance of R2 and R3 in parallel is less than R3 , so initially the current through R, is larger than its value after a long time has elapsed. ...
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 Spring '08
 Keil

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