# October 10, 2008 PHY2054 Discussion-Fall ‘08 Quiz 5 (Chapter 17.5-18.7) ```October 10, 2008
PHY2054 Discussion-Fall ‘08
Quiz 5 (Chapter 17.5-18.7)
Name:
UFID:
*1. (5pts) Suppose that you wish to fabricate a uniform wire out of 2.00 g of copper. If the wire is to
have a resistance of 0.300 Ω, and if all of the copper is to be used, what will be the length of the
wire? The resistivity and density of copper are 1.70&times;10-8 Ωm and 8.92&times;103 kg/m3.
The volume of the copper is
V = m/density = (2&times;10-3)/(8.92&times;103) = 2.24&times;10-7 m3
Since volume = cross sectional area &times; length, we can express resistance in terms of length and
volume:
R = ρL/A = ρL/(V/L) = ρL2/V ⇒ L = √(RV/ρ) = √[0.3&times;2.24&times;10-7/(1.70&times;10-8)] = 1.99 m
*2. (5pts) The heating coil of a hot water heater has a resistance of 8.00 Ω and it operates at 120 V.
Assuming that the water absorbs all of the energy converted by the heating coil, calculate how long
it takes to raise the temperature of 1.50 kg of water from 20.0 ˚C to the boiling point? (Specific heat
of water: C = 4.187&times;103 J/(kg˚C))
The power of the heating coil is
P = ΔV&sup2;/R = 120&sup2;/8 = 1800 w
The energy required to heat the water to the boiling point is
ΔW = mCΔT = 1.5&times;4.187&times;103&times;(100-20) = 5.02&times;105 J
Since power is the amount of energy supplied per unit time, we have
Δt = ΔW/P = 5.02&times;105/1800 = 279 s
**3. (5pts) The 8.0-V-emf battery has an internal resistance of 2.0 Ω, and the 9.0-V-emf battery has
an internal resistance of 3.0 Ω. What is the current in the 6.0-Ω resistor?
8.0 V
2.0 
We take I1 to be the current flowing in 2 Ω-resistor to the
right, I2 to be that flowing in 6 Ω-resistor to the right and I3
9.0 V
6.0 
b
a
to be that flowing in 4 Ω-resistor the left.
Applying Kirchhoff’s junction rule at point a, we have
I1+I2 = I3
4.0 
Applying Kirchhoff’s loop rule on the upper and lower loop
in the clockwise direction, we have
8-2I1-2I1+6I2+3I2-9 = 0 ⇒ 4I1-9I2 = -1 ⇒ I1 = (-1+9I2)/4
9-3I2-6I2-4I3 = 0 ⇒ 9I2+4I3 = 9 ⇒ I3 = (9-9I2)/4
Plugging I1 and I3 into the first equation yields
(-1+9I2)/4+I2 = (9-9I2)/4 = 0 ⇒-1+9I2+4I2-(9-9I2) = 0
⇒ 22I2 = 10 ⇒ I2 = 10/22 = 0.455 A to the right
***4. (5pts) Four capacitors are combined as shown. The combination is connected in series with a
4.00-kΩ resistor, a 120-V battery and a switch. What is the charge stored in the 24-μF capacitor 20.0
ms after the switch is closed?
12&micro;F 24&micro;F
20&micro;F
Combining C1 = 12 μF and C2 = 24 μF in series gives
1/C’ = 1/C1+1/C2 = 1/12+1/24 = 1/8 ⇒ C’ = 8 μF
Combining C’ and C3 = 12 μF in parallel gives
C” = C’+C3 = 8+12 = 20 μF
12&micro;F
C” and C4 = 20 μF are connected in series, thus the equivalent
capacitance of the combination is
1/C = 1/C”+1/C4 = 1/20+1/20 = 1/10 ⇒ C = 10 μF
The charge stored in the equivalent capacitor and the potential difference across it are
q(t) = CЄ(1-e-t/RC) &amp; ΔV(t) = q(t)/C = Є(1-e-t/RC)
C” and C4 are in series, thus they store the same charge. The potential difference across C” is
ΔV”(t) = q”(t)/C” = q(t)/C” = (C/C”)Є(1-e-t/RC)
C’ and C3 are in parallel, thus they have the same potential difference. Besides, C2 and C1 stores the
same charge since they are connected in series. Therefore,
q2(t) = q’(t) = C’ΔV’ = C’ΔV” = (C’C/C”)Є(1-e-t/RC)
At t = 20 ms, the charge in the 24 μF- capacitor is
q2(t = 20&times;10-3s) = (8&times;10/20)&times;120(1-exp[-20&times;10-3/(4&times;103&times;10&times;10-6)]) = 189 μC
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