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?

The ratio is .

Explanation:

We know ,  and  

Here all signs have their usual meanings.

Let us assume new voltage be and new length be .

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

     equation 1. ( Since , cross section area is same )

Same with power.

equation 2.

Diving equation 1 by 2.

We get ,

Putting this value of in equation 1.

We get,

Therefore,  

Hence, this is the required solution.

show the wavelengths of a wave. show the crests of a wave.

double the distance = divide strength by 4