A homogeneous wire is used to make a ring with a jumper on the diameter AB. What is the percentage change in the resistance between points A and B, if the jumper is cut?

###### Solution.

We denote the resistance of the upper and lower part of the ring as R_{1} and the resistance of the jumper R_{2}.

We start with a first case when we have the ring with a jumper.

The total resistance of the parallel circuit is determined using the following equation:

1/R_{total1} = 1/R_{1} + 1/R_{1} + 1/R_{2}

Then,

R_{total1} = (R_{1}*R_{2}/2)/(R_{1}/2+R_{2})

In the case when the jumper is cut, the total resistance of the parallel circuit is

1/R_{total} = 1/R_{1} + 1/R_{1}

or

R_{total2} = R_{1}/2

The circumference of the circle is given by the formula C = 2πr, where r is the radius of the circle.

Now, the resistance can be written by the resistance per unit length

R = λ*L, where λ is the resistance per unit length (Ω/m) and L is the length of the wire (m).

Then,

R_{1} = λ*πr (only half of the circle)

R_{2} = λ*2r (diameter)

Using this we will get

R_{total1} = 2λπr/(π+4)

and

R_{total2} = λπr/2

Finally, what percent of R_{total1} is R_{total2}?

R_{total2} is 178.5% of R_{total1}

The resistance increased by 78.5%