Answer:a). 97.2A b). 97.2A c). 5.85V/m d). 3.6V/m
e). 4.68V
Explanation:
Silver has a resistivity of 1.59e−8 Ω-m.
Copper has a resistivity of 1.68e−8 Ω-m.
Each wire is 0.60 mm in diameter or 0.30 mm in radius, giving a cross-sectional area of πr², or 2.83e-7 m² (note conversion from mm to m). Divide the resistivity by this number to get each wire's resistance per meter. Then multiply by the length of the wire to get the actual resistance:
Silver: (1.59e−8 Ω-m)(0.80 m) / (2.83e-7 m²) = 0.0450 Ω
Copper: (1.68e−8 Ω-m)(0.80 m) / (2.83e-7 m²) = 0.0475 Ω
Total resistivity: 0.0925 Ω (just the sum of the two wires)
Now that we know the resistance of each wire, we can easily answer the problems.
A) The current through the wires is found from Ohm's Law, V=IR. Solving for I gives:
I = V/R = (9.0 V) / (0.0925 Ω) = 97.2A
This same current passes through both wires, of course; any current passing through one wire must pass through the other.
B) I = 97.2A ( same as above)
C) Electric field strength E is measured in volts/meter. It is found by determining the voltage drop across a given length. For the 0.80-m long copper wire, the voltage drop is proportional to the copper wire's resistance. That is,
R(copper)/R(total) = V(copper)/V(total)
So
V(copper) = V(total) R(copper) / R(total)
= (9.0 V) (0.0475 Ω) / (0.0925 Ω)
= 4.68 V
Therefore, the total electric field E in the copper wire is
E(copper) = V(copper) / length(copper)
= (4.68V) / (0.80 m)
= 5.85 V/m
D) V(silver) = V(total) - V(copper)
= 9.0 V - 4.68 V
= 4.32 V
Thus,
E(silver) = V(silver) / length(silver)
= (4.32 V) / (1.2 m)
= 3.6 V/m
E) Already calculated in (C) =4.68V
