How is the voltage drop ΔV across the resistor related to the current I and the resistance R of the resistor? (Use any variable or symbol stated above as necessary.)

P = ΔV²/R

P = I²R

Explanation:

As a q passes by a resistor, it expends a (q V) where V is the potential drop between the resistor. This energy goes into radiation, much like the type a glob of putty that drops off a rock transforms its potential energy to heat when it strikes the bottom.

The transformation of potential energy into heat is referred to as dissipation. The power consumed in a resistor is the energy consumed per time. If an amount of Δq moves through the resistor in a time Δt, the power P = ΔqV/ Δt

                         P = ΔV²/R

                         P = I²R or IV

Where I is the current through the resistor, R and V is the voltage drop crossed it. The formula P = IV also provides the energy produced by a battery if I is the current spreading from the battery and V is its voltage.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

conduction, convection, radiation