Ohm’s Law, Electrical Power, Electrical Energy, Semiconductors, Superconductors Problems and Solutions 2
Problem #1
In Fig. 1, a battery of potential difference V = 12 V is connected to a resistive strip of resistance R = 6.0 Ω. When an electron moves through the strip from one end to the other, (a) in which direction in the figure does the electron move, (b) how much work is done on the electron by the electric field in the strip, and (c) how much energy is transferred to the thermal energy of the strip by the electron?
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| Fig.1 |
Answer;
Known:
Potential difference V = 12 V
Resistance R = 6.0 Ω
(a) in which direction in the figure does the electron move is from the positive pole to the negative pole
(b) work is done on the electron by the electric field in the strip,
W = Vit
With i = q/t, then
W = Vq = (12 V)(1.6 x 10─19 C) = 1.92 x 10─19 J
(c) energy is transferred to the thermal energy of the strip by the electron is
= i2R = (q/t)2
Problem #2
An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of 0.540 W. The same resistor is then connected between the terminals of a 1.50 V battery. At what rate is energy now dissipated?
Answer;
Known:
Potential difference battery (1) V1 = 3.00 V
Energy is dissipated in the resistor at the rate is P = 0.540 W
Potential difference battery (2) V2 = 1.50 V
because the power is proportional to the square voltage
P ∝ V2
rate is energy now dissipated is given by
P1/P2 = (V1/V2)2
0.540 W/P2 = (3.00 V/1.50 V)2
P2 = 0.135 W
Problem #3
A student kept his 9.0 V, 7.0 W radio turned on at full volume from 9:00 P.M. until 2:00 A.M. How much charge went through it?
Answer;
Known:
Potential difference V = 9.0 V
dissipation power is, P = 7.0 W
Time, t = from 9:00 P.M. until 2:00 A.M = 5 h = 18000 s
The dissipation power given by
P = Vi
i = P/V = 7.0 W/9.0 V = 0.78 A
Find the charge:
Q = it = 0.78 A x 18000 s = 14040 C = 14 kC
Problem #4
A 1250 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 1.0 h?
Answer;
Known:
dissipation power is, P = 1250 W
Potential difference V = 115 V
(a) the current in the heater when the unit is operating is
P = Vi, then,
i = P/V = 1250 W/115 V = 10.89 A
(b) the resistance of the heating coil is
R = V/i = 115 V/10.89 A = 10.58 Ω
(c) thermal energy is produced in 1.0 h is
W = Pt = 1250 W x 1.0 h = 1250.0 W = 1.250 kWh = 4.5 MJ
Problem #5
A copper wire of cross-sectional area 2.00 x 10─6 m2 and length 4.00 m has a current of 2.00 A uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in 30 min?
Answer;
Known:
A copper wire of cross-sectional area, A = 2.00 x 10─6 m2
Length, L = 4.00 m
Current, i = 2.00 A
(a) the magnitude of the electric field along the wire is
E = ρJ
With:
ρ = the resistivity of the copper wire = 1.68 x 10─8 Ωm
and J = current density = i/A
then,
E = ρ(i/A) = (1.68 x 10─8 Ωm)(2.00 A/2.00 x 10─6 m2) = 0.0168 V/m
(b) electrical energy is transferred to thermal energy in t = 30 min = 1800 s
W = Vit = ELit
W = (0.0168 V/m)(4.00 m)(2.00 A)(1800 s) = 241.92 J
Known:
Potential difference V = 12 V
Resistance R = 6.0 Ω
(a) in which direction in the figure does the electron move is from the positive pole to the negative pole
(b) work is done on the electron by the electric field in the strip,
W = Vit
With i = q/t, then
W = Vq = (12 V)(1.6 x 10─19 C) = 1.92 x 10─19 J
(c) energy is transferred to the thermal energy of the strip by the electron is
= i2R = (q/t)2
Problem #2
An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of 0.540 W. The same resistor is then connected between the terminals of a 1.50 V battery. At what rate is energy now dissipated?
Answer;
Known:
Potential difference battery (1) V1 = 3.00 V
Energy is dissipated in the resistor at the rate is P = 0.540 W
Potential difference battery (2) V2 = 1.50 V
because the power is proportional to the square voltage
P ∝ V2
rate is energy now dissipated is given by
P1/P2 = (V1/V2)2
0.540 W/P2 = (3.00 V/1.50 V)2
P2 = 0.135 W
Problem #3
A student kept his 9.0 V, 7.0 W radio turned on at full volume from 9:00 P.M. until 2:00 A.M. How much charge went through it?
Answer;
Known:
Potential difference V = 9.0 V
dissipation power is, P = 7.0 W
Time, t = from 9:00 P.M. until 2:00 A.M = 5 h = 18000 s
The dissipation power given by
P = Vi
i = P/V = 7.0 W/9.0 V = 0.78 A
Find the charge:
Q = it = 0.78 A x 18000 s = 14040 C = 14 kC
Problem #4
A 1250 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 1.0 h?
Answer;
Known:
dissipation power is, P = 1250 W
Potential difference V = 115 V
(a) the current in the heater when the unit is operating is
P = Vi, then,
i = P/V = 1250 W/115 V = 10.89 A
(b) the resistance of the heating coil is
R = V/i = 115 V/10.89 A = 10.58 Ω
(c) thermal energy is produced in 1.0 h is
W = Pt = 1250 W x 1.0 h = 1250.0 W = 1.250 kWh = 4.5 MJ
Problem #5
A copper wire of cross-sectional area 2.00 x 10─6 m2 and length 4.00 m has a current of 2.00 A uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in 30 min?
Answer;
Known:
A copper wire of cross-sectional area, A = 2.00 x 10─6 m2
Length, L = 4.00 m
Current, i = 2.00 A
(a) the magnitude of the electric field along the wire is
E = ρJ
With:
ρ = the resistivity of the copper wire = 1.68 x 10─8 Ωm
and J = current density = i/A
then,
E = ρ(i/A) = (1.68 x 10─8 Ωm)(2.00 A/2.00 x 10─6 m2) = 0.0168 V/m
(b) electrical energy is transferred to thermal energy in t = 30 min = 1800 s
W = Vit = ELit
W = (0.0168 V/m)(4.00 m)(2.00 A)(1800 s) = 241.92 J
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