A cylindrical metal wire at room temperature is carrying electric current betwee

Question

A cylindrical metal wire at room temperature is carrying electric current between its ends. One end is at potential VA = 50 V, and the other end is at potential VB = 0 V. Rank the following actions in terms of the change that each one separately would produce in the current from the greatest increase to the greatest decrease. In your ranking, note any cases of equality. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

(a) Make VA = 150 V with VB = 0 V.

(b) Adjust VA to triple the power with which the wire converts electrically transmitted energy into internal energy.

(c) Double the radius of the wire.

(d) Double the length of the wire.

(e) Double the Celsius temperature of the wire.

Explanation / Answer

c > a > b >e >d

explanation:

let initial current be i. Va = 50V
i1 = 50/R
a) when Va=150V, i' = 150/R = 3*i1
b)power = V2 / R,

power is tripled, so Va' = 3 Va

so i' = 3 *50/R = 3 * i1

c) R = *l/a where l is the length of wire, a is area = *r2 , r is the radius.

i = V/R = V*a/*l = V**r2 / *l

when radius is doubled, i becomes 4 times.

i' = 4*i1

d) from the above relation, when length is doubled, current is halved.

i' = (i1)/2

e) when temperature is increased resistance of the wire also increases but for small values the increase is small. R' = R(1+k) is the increase in temperature.

so here as temperature is doubled, R increases but it is < 2R as k is generally very small.

so i'<i1 but i'>i1/2.

hence the above relation.

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