213
HW Physics #5 Solutions
Spring 2009
26.23.
[Kirchoffs Circuit Rules] We have to use Kirchoffs rules to find the unknown currents, emfs, and resistances. 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 1, 2 and R. The circuit is sketched to the right. 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 b: 2.00 A I 4 3.00 A 0
I 4 1.00 A
Apply the junction rule to point c: I3 I 4 I5 0
I5 I3 I 4 8.00 A 1.00 A 7.00 A
As a check, apply the junction rule to point d: I5 2.00 A 5.00 A 0 (b) Apply the loop rule to loop (1): 1 3.00 A 4.00 I3 3.00 0
I5 7.00 A
1 12.0 V 8.00 A 3.00 36.0 V
Apply the loop rule to loop (2): 2 5.00 A 6.00 I3 3.00 0
2 30.0 V 8.00 A 3.00 54.0 V
(c) Apply the loop rule to loop (3): 2.00 A R 1 2 0
2 1 54.0 V 36.0 V 9.00 2.00 A 2.00 A Finally, we can apply the loop rule to loop (4) as a check of our calculations: 2.00 A R 3.00 A 4.00 5.00 A 6.00 0
R
2.00 A 9.00 12.0 V 30.0 V 0
18.0 V 18.0 V 0
26.50.
[Electric Dryer]
P 4100 W 17.1 A. Recall from page 201 that 12-gauge wire can carry up to 20A safely. V 240 V Thus, we can use 12-gauge wire safely for this electric dryer. V2 V 2 (240 V)2 14 . (b) P VI . Solving for the resistance, R R P 4100 W (c) At 11c per kWH, for 1 hour the cost is (11c/kWh)(1 h)(4.1 kW) 45c . (d) It makes sense that the cost to operate the device is proportional to its power consumption. The number also sounds about right. I am charged about one dollar to run a load of clothes through a dryer for an hour.
(a) P VI I 2 R , so I 23.82. [Alpha Particle Collision] As the alpha particle moves from a to b, its total energy, kinetic plus potential, is conserved: Ka U a Kb U b . Assume the particles initially are far apart, so U a 0 , The alpha particle has zero speed at the distance of closest approach, so K b 0 . Conservation of energy then gives K a U b =11MeV. U b
kq1q2 r k ( Z1e )( Z 2 e ) r
, where Z1 and Z2 are the atomic numbers (the
number of protons, each of which has charge +e). The alpha particle has Z1=2 and charge 2e while the lead nucleus has Z2=82 and charge 82e . Using this, and the fact that 1 eV 1.60 1019 J , we can get the radius: kq q k (2e)(82e) k (164)(1.60 1019 C)2 r 1 2 2.15 1014 m . Ub Ub (11.0 106 eV)(1.60 1019 J eV)
Its always a good idea to check that ones formula is reasonable. In this case, we see that the distance of closest approach becomes smaller if we throw the alpha particle at the lead nucleus with more initial energy. We also see that the distance would be larger if either particle had a larger charge. Both of these findings make intuitive sense, so it looks like our answer isnt crazy. By the way, since we treat both particles as point particles, this calculation assumes that at the distance of closest approach the alpha particle is outside the radius of the lead nucleus. If the alpha particle managed to penetrate into the lead nucleus, the problem would get more complicated. 23.22. [Pair of (+) charges] kq For a point charge, V . The total potential at any point is the algebraic sum of the potentials of the two charges. r (a) The positions of the two charges are shown at right. r a2 x 2 . (b) V0 2
1q . 4 0 a
(c) V (x) 2
a x2 (d) The graph of V versus x is sketched below.
2
1q 1 2 4 0 r 4 0
q
(e) When x a, V
1 2q , just like a point charge of charge 2q. At distances from the charges much greater than their 4 0 x
separation, the two charges act like a single point charge. (f) Ex(x) = dV(x) d 1 =- dx dx 4o 2q x +a
2 2
=
1 4 0
2 qx
x
2
a
2 3/2
, which is the expected electric field expression
E x(x) x
Lets check that our E graph and our V graph are consistent. We should have that Ex(x) = - slope of the V(x) vs. x graph. The slope of V is positive for x<0, zero at x=0, and negative for x>0. This is consistent with the electric field plot. 23.72.
b Since we know the electric field, we can get the potential from Va Vb E dl . By symmetry, the electric field is radially outward, a so E dl = E dr . Well integrate from a , so Va 0 .
From Example 22.9, we have the following. For r R : E
[Solid Insulating Sphere of Chrage]
kQ dr kQ , therefore V kQ 2 . In this formula, r is the radius at 2 r r r
r
which the potential is being computed, and to get I it used r as my integration variable. In this case, its not too important to keep them straight, but it can be in some problems.
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r R r kQ kQ r kQr kQ kQ 1 2 kQ kQ kQr 2 kQ r2 , therefore V E dr E dr 3 rdr 3 r 3 2 . 3 3 R R RR R R 2 R R 2R 2R 2R R R As a check, note that V is continuous at r=R.
For r R : E
(b) The graphs of V and E versus r are sketched below. E V
R
r
R
r
(c) There are a couple of ways to check that the V graph agrees with our formula. From the formulas, we see that V should decrease with r in both regions. V should be continuous at r=R. One can also easily check that dV/dr should be continuous at r=R. Also, dV/dr=0 at r=0. Finally, V should go to zero at infinity. The graph seems to have all of these properties. Q Q 23.73. Problem 23.72 shows that V for r R . (3 r 2 R 2 ) for r R and V 8 0 R 4 0 r Therefore V0 (a) V0 VR
3Q Q , VR 8 0 R 4 0 R
Q 8 0 R
(b) If Q 0 , V is higher at the center. If Q 0 ,V is higher at the surface. This makes sense. If the sphere is positively charged, it takes energy to push a positively charged particle from the surface to the center. If the sphere is negatively charged, a positively charge would be electrically attracted to the center.
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#1. [Current Divider] (a) Since R2 and R3 are in parallel, the voltage drop across one is equal to the voltage drop across the other. Let us call this common potential difference V23. Using Ohms law, V23 I 2 R2 I 3 R3 , so effective resistances. R2 and R3 are in parallel, so their effective resistance is R23 is Reff
I2 I3
R3 R2
. To actually compute the currents, we need to calculate
R2 R3 R2 R3
. R23 is in series with R1, so the total effective resistance
. The current through the battery is I R . The voltage across R1 is V1 IR1 . Then the eff voltage across R2, which is also the voltage across R3, is given by the loop rule (either loop which contains the battery):
R1 R23
R1R2 R1R3 R2 R3 R2 R3
V2 V3 V23 IR1
Then
V
R R R R2R3 R R 1 2 R1 3 2 3
R
3 2 I 2 R2 R R R R R R and I 3 R3 R R R R R R . 1 2 1 3 2 3 1 2 1 3 2 3 2 3
V
R
Once again, we find that
I2 I3
R3 R2
. Also, I 2 0 when R2 and I3 0 when R3 . . Let us compute this effective resistance and the currents through the resistors in two limits,
(b) As we found above, R23
R2 R3 R2 R3
using the formulae from part (a). When R3 , R23 R2 . I1 I 2
R1 R2
, I 0.
3
When R3 0, R23 0 . I1 I 3 , I 2 0 . R1 (c) Actually, we could have guessed all of the answers in part (b) without using the results of part (a). If R3 is infinite, no current will pass through it, so its just like removing the R 3 path from the circuit. Then R1 and R2 are effectively in series, and the currents are what you would expect. We could get the same effect by replacing R3 with an open switch, hence its name open-circuit. If R3=0, R3 is just like a zero-resistance piece of wire, so all of the current will pass through R 3 and none through R2. This is just like removing the R2 path from the circuit, so were left with just the battery and R1 (and R3). All of the current passes through R1 and R3, but theres no voltage drop across R3. The intuition to take away from this is that current likes to take the easiest (lowest resistance) path. #1. [Irreducible Circuit] (i) I = I1 + I2 (ii) (1) I5 = I2 I3
;
(iii)
(2) = = (3) I1R1 + I3R3 + I2R2 = I1R + I3R + 2I2R =0 Subtracting (1) from (2) gives I2 = I1 + I3. We find I1 3R
I1R1 I4R4 I2R2 I5R5
I4 = I3 + I1 =
;
I1R (I3 + I1) 2R 2I2R (I2 I3) R
=
3I1R 2I3R 3I2R + I3R
2 3 I 3 and
=0 =0
3R
I2
1 I3 . 3
(iv)
Substituting in (3), one finds I3 = /7R, so I1 = 3/7R and I2 = 2/7R. I3 has a negative value. That just means that the current actually flows in the direction opposite of the one drawn. (So you see that it doesnt matter whether your initial guesses for the current directions are accurate, as long as youre consistent. The initial choice really just decides what will count as positive current, and what will count as negative current.) (c) The current through the battery is I = I1 + I2 = 5/7R. (d) V3 = |I3R3| = |(/7R) R| = /7. The current, we found, goes from left to right. Since current always flows downstream, this means that the left end (point b) is at a higher potential than the right end (point c). The effective resistance between a and b must be Reff = /I = / (5/7R) = 7R/5 = (7/5) R.
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