A potential difference V is applied to a wire ofcross-section area of 1 unit, le

Question

A potential difference V is applied to a wire ofcross-section area of 1 unit, length 1 unit, and conductivity. You want to change the applied potential differenceand draw out the wire so the power dissipated is increased bya factor of 30 and the current is increased by a factorof 4 . What should be the new values of the length andthe new cross sectional area? A potential difference V is applied to a wire ofcross-section area of 1 unit, length 1 unit, and conductivity. You want to change the applied potential differenceand draw out the wire so the power dissipated is increased bya factor of 30 and the current is increased by a factorof 4 . What should be the new values of the length andthe new cross sectional area? A potential difference V is applied to a wire ofcross-section area of 1 unit, length 1 unit, and conductivity. You want to change the applied potential differenceand draw out the wire so the power dissipated is increased bya factor of 30 and the current is increased by a factorof 4 . What should be the new values of the length andthe new cross sectional area?

Explanation / Answer

the idea here is that the initial wire has resistance 1, andpotential V, so that using .     V = IR        wehave     V = I * 1 . Or     V = I    for thefirst wire .    In the final situation, youwant    I = 4V      becauseyou want the current to be four times as great . You also want the power to be 30 times as great. Initially,the power is VI which is the same as V2 . So now you want the power to be 30V2 . If we call V' the new voltage then we can write: .      V' = I'R'           V' =4V * R'          and     new power =V'I'        30V2 = V' 4V    . So... .              V' = 4VR'         V' =7.5V           7.5 V = 4V R' . or       R' = 7.5/4 = 1.875 . So original resistance is 1   and finalresistance is 1.875. . The idea is that     resistance = length / conductivity *area       and     volume = area*length . the volume of both wires is the same, since they are made fromthe same piece of metal. We can now write .            resistance = length / conduc *(volume/length)    or .     resistance = length2 /conduc * area . This means the resistance is proportional to the square of thelength, so the final length must be .       1.875 =      1.369units    . and the area is      1 / 1.369 =    0.7303units . If we call V' the new voltage then we can write: .      V' = I'R'           V' =4V * R'          and     new power =V'I'        30V2 = V' 4V    . So... .              V' = 4VR'         V' =7.5V           7.5 V = 4V R' . or       R' = 7.5/4 = 1.875 . So original resistance is 1   and finalresistance is 1.875. . The idea is that     resistance = length / conductivity *area       and     volume = area*length . the volume of both wires is the same, since they are made fromthe same piece of metal. We can now write .            resistance = length / conduc *(volume/length)    or .     resistance = length2 /conduc * area . This means the resistance is proportional to the square of thelength, so the final length must be .       1.875 =      1.369units    . and the area is      1 / 1.369 =    0.7303units

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