A potential difference V is applied to a wire of cross-sectional area A, length

Question

A potential difference V is applied to a wire of cross-sectional area A, length l, and conductivity . You want to change the applied potential difference and add length to the wire so that the power dissipated is increased by a factor of 30 and the current is increased by a factor of 4. Let the wire have the same cross-sectional area before and after it is lengthened.

(a) What is the ratio of the new length to the original length?

(b) What is the ratio of the new applied volrtage to the original voltage? [ Units V]

Explanation / Answer

I = VA/L

P = V2A/L

Call the new voltage V' and the new length L'

We want V'A/L' = 4VA/L and V'2A/L' = 30V2A/L

The and A drop out, write it as:

V'/L' = 4V/L and V'2/L' = V2/L, two simultaneous equations in unknowns V' and L'

Divide the second equation by the first to get: V' = (15/2)V which is the answer to part (b)

Plug that into the first equation to get L' = (15/8)L which is the answer to part (a)

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