Question

A potential difference V is applied to a wire of cross-section area A, length ℓ, and conductivity σ. 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. What is the ratio of the new length to the original length?

Answer 1

Answer:

The ratio is $\dfrac{15}{8}$.

Explanation:

We know , $I=\dfrac{V\sigma A}{L}$  and  $P=\dfrac{V^2\sigma A}{L}$

Here all signs have their usual meanings.

Let us assume new voltage be $V_1$ and new length be $L_1$.

Therefore, new value of I and P is compared by initial values.

$\dfrac{V_1\sigma A}{L_1}=4(\dfrac{V\sigma A}{L})\\\\\\\dfrac{V_1}{L_1}=\dfrac{4V}{L}$      equation 1. ( Since , cross section area is same )

Same with power.

$\dfrac{V_1^2\sigma A}{L_1}=30(\dfrac{V^2\sigma A}{L})\\\\\\\dfrac{V_1^2}{L_1}=30(\dfrac{V^2}{L})$ equation 2.

Diving equation 1 by 2.

We get , $V_1=\dfrac{15}{2}V$

Putting this value of $V_1$ in equation 1.

We get, $L_1=\dfrac{15}{8}L.$

Therefore,  $\dfrac{L_1}{L}=\dfrac{15}{8}$

Hence, this is the required solution.