A battery is used in an experiment and is connected by a 37.5 cm long gold wire with a diameter of 0.590 mm. If the battery has a voltage of 1.70 V and the resistivity of the wire is 2.4×10-8 Ω m, what is the current in the wire?
L = 0.375 m.
p = 2.4*10^-8 Ohm m.
r = 0.590/2 = 0.295 mm. = 2.95*10^-4 m.
A = pi*r^2 = 3.14*(2.95*10^-4)^2 = 2.7*10^-7 m^2.
R = pL/A = 2.4*10^-8*0.375/2.7*10^-7 = 0.033 Ohms.
I = V/R = 1.7/0.033 = 51A.
To find the current in the wire, we can use Ohm's Law, which states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to the resistance of the conductor. The resistance can be calculated using the formula:
Resistance (R) = (ρ x Length) / (Cross-sectional area)
Where:
- Resistance (R) is measured in ohms (Ω)
- Resistivity (ρ) is measured in ohm-meters (Ω*m)
- Length is the length of the wire in meters (m)
- Cross-sectional area is the area of the wire's cross-section in square meters (m^2)
Let's calculate the resistance of the wire first:
Given:
- Length of the wire (L) = 37.5 cm = 0.375 m
- Diameter of the wire (d) = 0.590 mm = 0.00059 m
- Resistivity of gold (ρ) = 2.4×10^-8 Ω*m
To calculate the cross-sectional area (A) of the wire, we can use the formula:
Cross-sectional area (A) = π x (diameter/2)^2
Now, we can plug these values into the formula:
A = π x (0.00059/2)^2
A ≈ 2.734 x 10^-7 m^2
Now, we can calculate the resistance:
R = (ρ x L) / A
R = (2.4×10^-8 x 0.375) / (2.734 x 10^-7)
R ≈ 3.291 Ω
Once we have the resistance, we can use Ohm's Law to find the current (I):
I = V / R
Given:
- Voltage (V) = 1.70 V
- Resistance (R) = 3.291 Ω
I = 1.70 / 3.291
I ≈ 0.516 A
Therefore, the current flowing through the wire is approximately 0.516 Amperes.