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Unformatted text preview: "10.3 _
a) The speed of each proton can be found from its kinetic energy. Since we want the speed in units of
meters per second, ﬁrst convert the kinetic energy from electron—volts to joules, 3 #19 
1 2 “lg: W24_38x105m/3_
m KB = im" :1" 1" Z 1.67 x 10'27 kg b) Using Equation 19.3 on page 840 of the text, the current I = nqA(v). The cross—sectional area of the
beam is A = m2 = n(1.00 x 10—3 m)2 z 3.14 x 10*5 m 2.
The number n of charge carriers per cubic meter is 102 cm 1m 3
n = 1.20 X 106 protons/cm3 ( > = 1.20 x 1012 protons/n13. Hence, using Equation 19.3 on page 840 of the text, the current is I = nqA('u)
2:» r z (1.20 x 1012 protons/m3 )(1.602 x 10—19 C/proton )(3.14 x 106 m2)(4.38 x 105 m/s) =31: 2.64 x 10—7 A = 0.264 “A. “19.? Since the—charge is uniformly distributed over the perimeter, the charge on each side of the Square
is (3/4. The side along the rotation axis contributes nothing to the current since it is not in motion. The
other three sides pass the line once each revolution, contributing a charge 3Q/4 to the current. The current is the number of coulombs per second passing any given point along the cirCumference, which is just the charge times the frequency. The frequency of the motion is w =é—7F‘ and therefore, 1:. (%) if? 19.24
a.) The Current in the wire is the same at all points along its length, so the current in the thin section also is 20.0 A.
b) The resistance of the 3.00 In length of wire is given by Equation 19.13 on page 842 of the text, p3 (1.77 x 10—8 Qrn)(3.00 In) _3
= — = H = 7.5 (2.
R3 A imam x 10—3 [in/212 1 X 10
c) The resistance of the 1.00 m length of wire is _ pf _ (1.77 x 10—3 Qrn )[100 In) ﬂ _3
““ A " 7r[(1.00 x 103 rim/212 ‘ 225 X 10 9' d) The two resistances are in series, so the total resistance is their sum,
R: R3 +R1 = 7.51x 103 a + 22.5 x 103 n = 30.0 x 10—3 9. R1 9) The potential difference between the ends of the wire is found from Ohm’s law,
V = IR = (20.0 A)(30.0 'x 103 n) = 0.600 v. 19.26 The wire connecting the top and bottom makes this junction one large node. Redraw the circuit,
progressing from A to B: 59 100 109 59 The two pairs of 10 ﬂ resistors are in parallel, so the equivalent resistor has a resistance equal to their
product divided by the sum, (10 man a) ioo+1on 25'09' Using this result, our circuit looks like 59 5.051 5.09 59 The four remaining resistors are in series, so their equivalent resistor has resistance 59+5.0Q+5.09+59 =2OQ. 19.59 a) The battery and 15.0 9 resistor are connected to the same two nodes, and so are connected in parallel.
Therefore, they have the same potential difference across them. Hence the potential diﬁerence across the
15.0 9 resistor is 30.0 V. ' b) The power absorbed by the 15.0 9 resistor is
P = IV,
From part a) we know that V = 30.0 V, and we can use Ohm‘s law to ﬁnd the current, since
V=IR => 30.0V =I(15.0Q) =,* I=2.00A. Therefore, the power absorbed by the 15.0 Q resistor is P 2 IV = (2.00 A)(30.0 V) = 60.0 W.
c) The two 10.0 9 resistors are in series, so the resistance of their equivalent resistor is 10.0 9 + 10.0 9 = 20.0 Q. The arrangement then looks iike: ® 15.0 Q 20.0 Q The 20.0 Q _resistor is in parallel with the source; hence the potential difference across it also is 30.0 V,
and we can use Ohm’s law to ﬁnd the current through it. V=IR : 30.0V =r(20.09) => 1721.50.41. d) To answer this question we need to know the current passing through the battery. Draw a diagram showing everything we know about the circuit and the apply Kirchhoff ’s current law to the node labeled A
in the sketch below to ﬁnd the current I’ through the battery: 200 Q Let I’ be the current through the battery. Then, using the KCL at node A,
—I’+2.00A +1.50A =0A =} I’ =3.50A. The power absorbed by the battery is the product of the current into its positive polarity terminal and the
potential difference across it, P = (—3.50 A)(30.0 V) = —105 w. e) The current through each of the 10.0 Q resistors is 1.50 A . The potential difference across each is found
from Ohm’s law; for either resistor, V 2 IR: (1.50A)(10.0 Q) = 150V.
The power absorbed by each is P = IV = (1.50 A)(15.0 V) = 22.5 W. 10.51
a) The power absorbed by the fuse is
P = IV. Use Ohm’s law to substitute for V, and then solve for I,
b) P 30W
_ = 2 = —: =1A.
P—I{IR} rn=>r MR 003“) 0
c) and d) Since all the circuit elements are in series, the current is the same at points A and B. Since all the
resistors are in series, the resistance of the equivalent resistor is seq; 0.309 +1.09 +13.09 +1.09 = 15.3 9, and the equivalent circuit is just the independent voltage source in series with the equivalent resistor. To
ﬁnd the current I, we can apply the KVL cloekwise around the loop, 120 V
15.3 Q Therefore, the current at all points along the original circuit is 7.84 A.
e) The potential difference across the heater is feund from Ohm ’s law, V = rs = [7.84 A)(13.0 9) z 102 V. —120V+IReq=0V =>~I= =7.84A. f) H ‘Beééﬁ's'e the two hEaters are in parallel, they are equivalent to' one—resistor with _ (13.0 Q )(13.0 9)
3“ 13.0 9 + 13.0 Q Now proceed just as we did above. Since all of the resistors are in series, the total resistance in the circuit is
Ram eq = 0.3!) Q + 1.0 Q + 5.50 Q. + LO 9 2 8.8 Q, and the equivalent circuit has a resistor with this resistance in series with the power source. Hence, we can
apply the KVL clockwise around the loop to ﬁnd the current, ' = 6.50 Q. 120V
m—14A. This exceeds the capacity of the fuse (calculated in part a)) and the fuse will melt. 120V+IR€Q=UV :1» I: 19.68
a) The potential energyr stored in the capacitor is m: gov? :1» 1.00 .1 = $00500 V)2 => 0 = 8.00 X 106 F = 8.00pF. b) The time constant is r = R0, so five time constants is 57 = 5R0 = 0.100 x 10—3 s 2 51:18.00 x 10—6 F) => R = 2.50 Q. C) From Example 19.19 on page 875, the potential energy varies with time as me) = Page2W. The instantaneous power P is dPE 2 _ T 2 = E'— = —;1:E'0€ 2t; 2 The average power over the 5’?" time interval is . 1 5T 1 “are
(P) 51' 05 P£) 57 OS alt
PEST PEG 10 m,  FED
=—u =_—  1 =——.r.1.:1 rh1=——
5'1" Us 51"(e ) 5T ( 5 XlO ) 51"(1) = —1§£‘1—— — 4.00 x 103w. _5(0.100 x 103 s) ” So, as the capacitor bank discharges, it supplies an average power of 2.00 kW, or 2000 J / s. 19.7 2
a) The current in the branch containing the capacitor is zero, since the capacitor acts as an Open switch. b) Since there is no current in the branch with the 1.0 kg resistor, the potential difference across it is 0 V
(from Ohm’s law). c) The circuit effectively is only a single loop, since the part between nodes A and B with the capacitor
carries no current. Furthermore, the resistors in the part of the circuit carrying current are in series, so the}r
can be replaced by a Single resistor with resistance 2.0 k9 + 3.0 M? = 5.0 k9. To ﬁnd the current, apply the KVL clockwise around the 1001),
—20 V +I(5.0 x 103 9) 29V => I: 4.0 x 10—3 A =4.0n1A._ d) Use Ohm‘s law:
16.0 m = IR: (4.0 x 10‘3 A)(3.0 x 103 Q} = 12 V_ e) Since there is ,0 V across the 1.0 k9 resistor, the capacitor effectively is in parallel with the 3.0 kQ
resistor, and so also has 12 V across it. E) There is zero power absorbed by the capacitor and the 1.0 k9 resistor. The power absorbed by the
3.0 k9 resistor is Psnkn = “3.0m = (40 X 10—3 A)(12V) = 4.8 X 10—2 W.
The power absorbed by the 2.0 it!) resistor is
PM W = IVZAU mi = IUR) = Pa = (4.0 x 103 A)2(2.0 x 103 Q) = 3.2 x 102 W.
The power absorbed by the battery is
P20 v = (—4.0 x 103 A J(20 V) = —8.0 x 102 w, The total power absorbed by all circuit elements is Ptota12—80x10‘zw +3.2x 102w +4.8x10‘2W+0W +0w =0W. ...
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This note was uploaded on 10/12/2008 for the course PHYSICS EN.171.102 taught by Professor Barnett during the Fall '07 term at Johns Hopkins.
 Fall '07
 Barnett
 Physics

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