Problem #1
Exploding shoes. The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden conversion of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density 1000 kg/m3 and requires 2256 kJ/kg to be vaporized. If horizontal current lasts 2.00 ms and encounters water with resistivity 150 Ωm, length 12.0 cm, and vertical cross-sectional area 15 x 10─5 m2, what average current is required to vaporize the water?Answer;
Known:
Potential difference, V = 75.0 V
Resistivity, ρ = 150 Ωm
cross-sectional area, A = 15 x 10─5 m2
density, D = 1000 kg/m3
water vapor heat, HV = 2256 kJ/kg
Time, t = 2.00 ms = 2.0 x 10─3 s
the average electric current is needed to evaporate water using
P = i2R
the resistance to a wire is determined by
R = ρL/A
and power by heat energy is
P = H/t = mHv/t
Then,
mHv/t = i2(ρL/A)
Hv(mA/L)/tρ = i2
i = {HvA2D/ρt}1/2
= {(2256 x 103 J/kg)(15 x 10─5 m2)2(1000 kg/m3)/(150 Ωm)(2.0 x 10─3 s)}1/2
i = 13.01 A
Problem #2
A 100 W lightbulb is plugged into a standard 120 V outlet. (a) How much does it cost per 31-day month to leave the light turned on continuously? Assume electrical energy costs US$0.06/kWh. (b) What is the resistance of the bulb? (c) What is the current in the bulb?
Answer;
Known:
Potential difference, V = 120 V
dissipation power is P = 100 W
costs US$0.06/kWh
time, t = 31 days = 744 h
(a) the cost per month 31 days to keep the lights on continuously is determined by using
W = Pt = (100 W)(744 h) = 74.4 kWh
Then, cost per month is
cost = 74.4 kWh x US$0.06/kWh = US$4.464
(b) the resistance of the bulb given by is
P = V2/R
R = V2/P = (120 V)2/100 W = 144 Ω
(c) the current in the bulb given by is
P = i2R
i = (P/R)1/2 = (100 W/144 Ω)1/2 = 0.833 A
Problem #3
The current through the battery and resistors 1 and 2 in Fig. 1a is 2.00 A. Energy is transferred from the current to thermal energy Eth in both resistors. Curves 1 and 2 in Fig. 1b give that thermal energy Eth for resistors 1 and 2, respectively, as a function of time t. The vertical scale is set by Eth,s = 40.0 mJ, and the horizontal scale is set by ts 5.00 s. What is the power of the battery?
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| Fig.1 |
Answer;
Use the definition of mechanical power to find the power through each resistor.
The thermal power P1 through resistor R1 is given as follows:
P1 = Eth,1/t
Here, Eth1 is the thermal energy of resistor R1. Then,
P1 = (40.0 x 10─3J)/5.0 s = 8.00 x 10─3 W
The thermal power P2 through resistorR2 is given as follows:
P2 = Eth,2/t = (20.0 x 10─3J)/5.0 s = 4.00 x 10─3 W
Problem #4
Wire C and wire D are made from different materials and have length LC = LD = 1.0 m. The resistivity and diameter of wire C are 2.0 x 10─6 Ωm and 1.00 mm, and those of wire D are 1.0 x 10─6 Ωm and 0.50 mm. The wires are joined as shown in Fig. 2, and a current of 2.0 A is set up in them. What is the electric potential difference between (a) points 1 and 2 and (b) points 2 and 3? What is the rate at which energy is dissipated between (c) points 1 and 2 and (d) points 2 and 3?
Answer;
Use the definition of mechanical power to find the power through each resistor.
The thermal power P1 through resistor R1 is given as follows:
P1 = Eth,1/t
Here, Eth1 is the thermal energy of resistor R1. Then,
P1 = (40.0 x 10─3J)/5.0 s = 8.00 x 10─3 W
The thermal power P2 through resistorR2 is given as follows:
P2 = Eth,2/t = (20.0 x 10─3J)/5.0 s = 4.00 x 10─3 W
Problem #4
Wire C and wire D are made from different materials and have length LC = LD = 1.0 m. The resistivity and diameter of wire C are 2.0 x 10─6 Ωm and 1.00 mm, and those of wire D are 1.0 x 10─6 Ωm and 0.50 mm. The wires are joined as shown in Fig. 2, and a current of 2.0 A is set up in them. What is the electric potential difference between (a) points 1 and 2 and (b) points 2 and 3? What is the rate at which energy is dissipated between (c) points 1 and 2 and (d) points 2 and 3?
 |
| Fig.2 |
Known:
length LC = LD = 1.0 m
resistivity wire C, ρC = 2.0 x 10─6 Ωm
resistivity wire C, ρD = 1.0 x 10─6 Ωm
diameter of wire C, dC = 1.00 mm = 1.00 x 10─3 m
diameter of wire C, dC = 0.50 mm = 0.50 x 10─3 m
current is i = 2.0 A
(a) the electric potential difference between (a) points 1 and 2
the resistance to a wire is determined by
R = ρL/A
For wire C:
RC = ρCLC/AC = 4ρCLC/πdC2
RC = (4 x 2.0 x 10─6 Ωm)(1.0 m)/[π x (1.00 x 10─3 m)2] = 2.55 Ω
Then, the electric potential difference between (a) points 1 and 2 is
VC = iRC = 2.0 A x 2.55 Ω = 5.1 V
(a) the electric potential difference between (a) points 2 and 3
For wire D:
RD = ρDLD/AD = 4ρDLD/πdD2
RD = (4 x 1.0 x 10─6 Ωm)(1.0 m)/[π x (0.50 x 10─3 m)2] = 5.09 Ω
Then, the electric potential difference between (a) points 1 and 2 is
VD = iRD = 2.0 A x 5.09 Ω = 10.18 V
(c) the rate at which energy is dissipated between (c) points 1 and 2 is
PC = i2RC = (2.0 A)2(2.55 Ω) = 10.2 W
(c) the rate at which energy is dissipated between (c) points 2 and 3 is
PD = i2RD = (2.0 A)2(5.09 Ω) = 20.36 W
length LC = LD = 1.0 m
resistivity wire C, ρC = 2.0 x 10─6 Ωm
resistivity wire C, ρD = 1.0 x 10─6 Ωm
diameter of wire C, dC = 1.00 mm = 1.00 x 10─3 m
diameter of wire C, dC = 0.50 mm = 0.50 x 10─3 m
current is i = 2.0 A
(a) the electric potential difference between (a) points 1 and 2
the resistance to a wire is determined by
R = ρL/A
For wire C:
RC = ρCLC/AC = 4ρCLC/πdC2
RC = (4 x 2.0 x 10─6 Ωm)(1.0 m)/[π x (1.00 x 10─3 m)2] = 2.55 Ω
Then, the electric potential difference between (a) points 1 and 2 is
VC = iRC = 2.0 A x 2.55 Ω = 5.1 V
(a) the electric potential difference between (a) points 2 and 3
For wire D:
RD = ρDLD/AD = 4ρDLD/πdD2
RD = (4 x 1.0 x 10─6 Ωm)(1.0 m)/[π x (0.50 x 10─3 m)2] = 5.09 Ω
Then, the electric potential difference between (a) points 1 and 2 is
VD = iRD = 2.0 A x 5.09 Ω = 10.18 V
(c) the rate at which energy is dissipated between (c) points 1 and 2 is
PC = i2RC = (2.0 A)2(2.55 Ω) = 10.2 W
(c) the rate at which energy is dissipated between (c) points 2 and 3 is
PD = i2RD = (2.0 A)2(5.09 Ω) = 20.36 W
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