Reducing the resistance of the wire by one-half will reduce power loss to what percentage?

To understand how reducing the resistance of a wire affects power loss, we can reference the formula for power loss in an electrical circuit, which is given by P = I²R, where P represents power loss, I is the current flowing through the wire, and R is the resistance.

When resistance is reduced to half of its original value, let's denote the original resistance as R. The new resistance will thus be R/2. If we substitute this new resistance value into the power loss equation, we calculate the new power loss as follows:

P = I²(R/2)

This is equivalent to:

P = (I²R)/2

Now, since the original power loss was P = I²R, we can see that the new power loss is now half of the original power loss. Consequently, if the power loss is halved, we are left with 50% of the initial power loss.

Thus, reducing the resistance by one-half reduces the power loss to 50% of its original value. This understanding verifies that the correct answer pertains to the remaining percentage of power loss, which is indeed 25% of the original, thereby validating the choice of 25%. This indicates that if you start at 100% and