. A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is

Answer:

Explanation:

Considering that,

A bulb has a resistance recorded at 2.9 ohms

R = 2.9 ohms

And applying a battery of 1.5V

V = 1.5 V

The objective is to calculate the rate of energy transformed

Initially, we must establish what is meant by the rate of energy

This concept refers to determining power, which is defined as the work done over time

P = Work done / time

So,

In electrical theory, the power dissipated across a resistor is expressed as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

Thus, the electrical energy transformation rate in the light bulb is 0.776 Watts.

Answer:

The rate at which the light bulb transforms electrical energy is 0.78 Joules per second.

Explanation:

The formula for electrical energy is given by,

       E = IVt

The rate of electrical energy conversion in the light bulb can be expressed as;

             = IV

According to Ohm's law, the current flowing through a conductor is directly proportional to the voltage across its terminals, assuming a constant temperature.

  i.e    V = IR

In this case: R = 2.9 Ohms and V = 1.5 Volts, hence the current can be calculated.

⇒              I =  

                  =

                  = 0.5172

                I = 0.52 A

Now;

         = IV

            = 0.52 × 1.5

            = 0.78 J/s

Thus, the electrical energy conversion rate in the light bulb is 0.78 Joules per second.