Respuesta :
Answer:
Therefore,
Current through Nichrome wire is 0.3879 Ampere.
Explanation:
Given:
Length = l = 10 meter
[tex]Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter[/tex]
[tex]Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter[/tex]
V = 12 Volt
To Find:
Current, I =?
Solution:
Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as
[tex]R=\dfrac{\rho\times l}{A}[/tex]
Where,
R = Resistance
l = length
A = Area of cross section = πr²
[tex]\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter[/tex]
Substituting the values we get
[tex]R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}[/tex]
[tex]R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}[/tex]
[tex]R=\dfrac{1\times 10^{2}}{3.23}[/tex]
[tex]R=30.95\ ohm[/tex]
Now by Ohm's Law,
[tex]V= I\times R[/tex]
Substituting the values we get
[tex]I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere[/tex]
Therefore,
Current through Nichrome wire is 0.3879 Ampere.
The electric current is defined s the rate of flow of charge in the wire. The current through the wire will be 0.3786 A.
What is electric current?
The electric current is defined s the rate of flow of charge in the wire. The current through the wire . Its unit is ampere(A). It is denoted by the letter I.
The given data in the problem is;
l is the length of the wire = 10 m
r is the radius of the wire= 0.321 mm = 0.321 ×10⁻³
V is the voltage of the battery = 12.0-V
[tex]\rm \rho[/tex] is the resistivity of the nichrome wire= 1.00 × 10-6 Ω ∙ m.
The resistance of the wire is found by the formula;
[tex]\rm R= \frac{\rho L}{A} \\\\ \rm R= \frac{1\times 10^{-6}\times 10}{3.23} \\\\ \rm R= 30.95 \ ohm[/tex]
According to ohms law, The current is the ratio of voltage and resistance;
[tex]\rm I = \frac{V}{R} \\\\ \rm I = \frac{12}{30.95} \\\\ \rm I =0.3876\ A[/tex]
Hence the current through the wire will be 0.3786 A.
To learn more about the electric current refer to the link;
https://brainly.in/question/3122013
