how do u find area of cross section when resistance resistivity and length of current carrying conductor is given using formula r=rho l/a? ans fast if ans is appropriate i will mark it as brainliest
Answers
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Answer:
HEY MATE HERE'S UR ANSWER
Explanation:
PHYSICS
The resistance R in terms of length Land cross section A and resistivity ρ of wire is given by
A .
A2ρL
B .
AρL
C .
A2ρL2
D .
AρL2
ANSWER
The total amount of resistance to charge flow within a wire of an electric circuit is affected by some clearly identifiable variables.
The total length of the wires will affect the amount of resistance. The longer the wire, the more resistance that there will be. There is a direct relationship between the amount of resistance encountered by charge and the length of wire it must traverse. If resistance occurs as the result of collisions between charge carriers and the atoms of the wire, then there is likely to be more collisions in a longer wire. More collisions mean more resistance.
The cross-sectional area of the wires will affect the amount of resistance. Wider wires have a greater cross-sectional area. The wider the wire, the less resistance that there will be to the flow of electric charge. When all other variables are the same, charge will flow at higher rates through wider wires with greater cross-sectional areas than through thinner wires.
✔FURTHER EXPLANATION✔
Here,
\rhoρ = resistance
R = resistance
A = cross-sectional
L = length
Therefore, resistance \propto∝ \frac{1}{A}A1 of cross section of the conductor (A)
Or, R \propto∝ \frac{1}{A}A1 (2)
Now from equation (1) and (2)
R \propto∝ \frac{l}{A}Al
Or, R = p\frac{l}{A}pAl (3)
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Answer:
hey mate here's ur answer
Explanation:
The resistance R in terms of length Land cross section A and resistivity ρ of wire is given by
A .
A2ρL
B .
AρL
C .
A2ρL2
D .
AρL2
ANSWER
The total amount of resistance to charge flow within a wire of an electric circuit is affected by some clearly identifiable variables.
Here,
\rhoρ = resistance
R = resistance
A = cross-sectional
L = length
Therefore, resistance \propto∝ \frac{1}{A}A1 of cross section of the conductor (A)
Or, R \propto∝ \frac{1}{A}A1 (2)
Now from equation (1) and (2)
R \propto∝ \frac{l}{A}Al
Or, R = p\frac{l}{A}pAl (3)
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