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Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
Answer:
5.33*10^-3 seconds
Explanation:
c = d/t
c = speed of light constant (3.0*10^5 km/s)
d = distance (1600 km)
t = ?
3.0*10^5 = 1600/t
t = 1600/3.0*10^5
t = 5.33*10^-3 seconds
I hope this helped! :)
Answer:
The current will be increased and also for the resistance.
Explanation:
The analysis of a direct current circuit can give us the explanation we need. Using the ohm law, which tells us that the voltage is equal to the product of the current by the resistance we have:
The voltage is equal to the potential difference therefore we will have these expressions:
If we increase the potential differential or circuit voltage, the current will also increase and so does the resistance by increasing the voltage. If we put numerical values in the equation given before, we can confirm this fact.