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 resistivity ?. You want to change the applied                     potential difference and stretch the wire so that the energy dissipation rate is multiplied by 30.0 and the current is multiplied by 4.24. Assuming the wire's                     density does not change, what are (a) the ratio of the new length to L and (b) the ratio of the new cross-sectional area to A?                     Please show work and type out answer

Explanation / Answer

energy dissipation rate = R = VI

R2/R1 = 30

I2/I1 = 4.24

therfore V2/V1 = 30 /4.24 = 7.0755

I = V/r where r is resistance

I2/I1 = (V2/V1 ) x (r1/r2)

4.24 = 7.0755 x (r1/r2)

r1/r2 = 0.6

r = (resistivity) x L /A

as the original wire is streched so the volume is same

L1A1 = L2 A2

A2/A1 = L1/L2

r1/r2 =( L1/L2 ) x (A2/A1) =(L1/L2)^2 = 0.599

L1/L2 = 0.774

L1 = L and A1 = A

L2/L = 1.2918

A2/A = 0.774

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