HOMEWORK 4 4. Let Q1 and V1 be the initial charge and voltage on the capacitor, and let Q2 and V2 be the final charge and voltage on the capacitor. Use Eq. 24-­‐1 to relate the charges and voltages to the capacitance. Q1 = CV1 Q2 = CV2 Q2 − Q1 = CV2 − CV1 = C V2 − V1 →
(
5. C=
Q2 − Q1
V2 − V1
26 × 10 C
=
50 V
= 5.2 × 10 −7 F = 0.52 µ F
After the first capacitor is disconnected from the battery, the total charge must remain constant. The voltage across each capacitor must be the same when they are connected together, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-­‐potential connecting wire. Use the total charge and the final potential difference to find the value of the second capacitor. QTotal = C1V1
Q1
initial
)
−6
QTotal = Q1
+ Q2
final
= C1Vfinal
Q2
final
= C2Vfinal
final
(
)
= C1 + C2 Vfinal → C1V1
final
(
)
= C1 + C2 Vfinal → initial
⎛ V1
⎞
⎛ 125V ⎞
initial
C2 = C1 ⎜
− 1⎟ = 7.7 × 10 −6 F ⎜
− 1⎟ = 5.6 × 10 −5 F = 56 µ F
⎜ Vfinal
⎟
⎝ 15V
⎠
⎝
⎠
(
)
23. We want a small voltage drop across C1. Since V = Q C , if we put the smallest capacitor in series with the battery, there will be a large voltage drop across it. Then put the two larger capacitors C3
in parallel, so that their equivalent capacitance is large and therefore will have a small voltage drop across them. So put C1 V0
C2
and C3 in parallel with each other, and then put that combination in series with C2. See the diagram. To calculate the voltage across C1, find the equivalent capacitance and the net charge. That charge is used to find the voltage drop across C2, and then that voltage is subtracted from the battery voltage to find the voltage across the parallel combination. Qeq
C C + C3
C + C 2 + C3
Q
1
1
1
=
+
= 1
→ Ceq = 2 1
; Qeq = CeqV0 ; V2 = 2 =
;
Ceq C2 C1 + C3 C2 C1 + C3
C1 + C2 + C3
C2 C2
(
(
)
(
C2 C1 + C3
V1 = V0 − V2 = V0 −
Qeq
C2
= V0 −
)
CeqV0
C2
= V0 −
)V
C1 + C2 + C3
C2
0
=
C2
C1 + C2 + C3
V0 =
1.5µ F
6.5µ F
(12 V )
= 2.8 V
30. C1 and C2 are in series, so they both have the same charge. We then use that charge to find the voltage across each of C1 and C2. Then their combined voltage is the voltage across C3. The voltage across C3 is used to find the charge on C3. C1
Q1
Q1 = Q2 = 12.4 µC ; V1 =
C1
=
12.4 µC
16.0µ F
= 0.775V ; V2 =
(
)(
Q2
C2
=
12.4 µC
16.0µ F
= 0.775V
)
V3 = V1 + V2 = 1.55V ; Q3 = C3V3 = 16.0 µ F 1.55V = 24.8µC
From the diagram, C4 must have the same charge as the sum of the charges on C1 and C3. Then the voltage across the entire combination is the sum of the voltages across C4 and C3. Q
37.2 µC
Q4 = Q1 + Q3 = 12.4 µC + 24.8µC = 37.2 µC ; V4 = 4 =
= 1.31V
C4 28.5µ F
Vab = V4 + V3 = 1.31V + 1.55V = 2.86 V
Here is a summary of all results. Q1 = Q2 = 12.4 µC ; Q3 = 24.8µC ; Q4 = 37.2 µC
V1 = V2 = 0.775V ; V3 = 1.55V ; V4 = 1.31V ; Vab = 2.86 V
5. (a) (b) )(
(
18. Use Eq. 25-­‐2b to find the current. V 240 V
V = IR → I = =
= 27.91A ≈ 28A R 8.6 Ω
Use the definition of current, Eq. 25-­‐1a. ΔQ
I=
→ ΔQ = I Δt = 27.91A 50 min 60s min = 8.4 × 10 4 C Δt
)(
)
Use Eq. 25-­‐5 for the resistivity. )
(
ρT Al = ρ0 Al ⎡⎣1 + α Al T − T0 ⎤⎦ = ρ0 W →
T = T0 +
21. 1 ⎛ ρ0 W
α Al ⎜⎝ ρ0 Al
1
⎛ 5.6 × 10 −8 Ωim
⎞
− 1⎟ = 279.49° C ≈ 280° C
−1 ⎜
−8
0.00429 ( C° ) ⎝ 2.65 × 10 Ωim ⎠
⎞
− 1⎟ = 20° C +
⎠
The wires have the same resistance and the same resistivity. Rlong = Rshort →
ρl long
A1
=
ρl short
A2
→
( 4) 2l
2
short
π d long
=
4l short
2
π dshort
→
d long
dshort
=
2 36. (a) Since P =
V2
R
→ R=
V2
P
says that the resistance is inversely proportional to the power for a constant voltage, we predict that the 850 W setting has the higher resistance. (b) R=
V2
P
(120 V )
=
850 W
2
= 17 Ω (c) R=
V2
P
(120 V )
=
2
1250 W
= 12 Ω