91Ó°ÊÓ

The time the bulbs are on for 3.00 yr is

$$\begin{gathered} t=\left(4.0h/day\right)\left(3.00yr\right)\left(\frac{365days}{1.00yr}\right) \\ t=4380h \end{gathered}$$

For the compact fluorescent bulbs:

Electrical energy consumed in 3.00 yr

$$\begin{gathered} E_f=P_{f}t \\ {E}_f=\left(23.0W\right)\left(4380h\right) \\ {E}_f=101KW-h \end{gathered}$$

Cost for this electrical consumption

$$\begin{gathered} C_{fe}=E_{f}S=\left(101KW-h\right)\left(0.080\$/KW-h\right) \\ {C}_{fe}=\$8.06 \end{gathered}$$

Thus, the total cost is

$$\begin{gathered} C_f=C_{fe}+C_{fb}=\$8.06+\$11.0 \\ {C}_f=\$19.1 \end{gathered}$$

For the incandescent bulbs:

Electrical energy consumed in 3.00 yr

$$\begin{gathered} E_i=P_{i}t=\left(100W\right)\left(4380h\right) \\ {E}_i=438KW-h \end{gathered}$$

Cost of this electrical consumption:

$$\begin{gathered} C_{ie}=E_{i}S=\left(438KW-h\right)\left(0.080\$/KW-h\right) \\ {C}_{ie}=\$35.0 \end{gathered}$$

Thus, the total cost is:

$$\begin{gathered} C_i=C_{ie}+C_{ib}=\$35.0+\$0.75 \\ {C}_i=\$35.8 \end{gathered}$$

Now, subtract the cost of fluorescent bulbs from the cost of incandescent bulbs to obtain the total savings

$$\begin{gathered} S=C_i-C_f \\ S=\$35.8-\$19.1 \\ S=\$16.7 \end{gathered}$$

Therefore, the total savings for over the 3.00 yr is $ 16.7

Use the formula of power in relation with potential difference and resistance

$$\begin{gathered} P=\frac{V^2}{R} \\ R=\frac{{\left(120V\right)}^2}{23.0W} \\ R=626\mathrm{Î\copyright } \end{gathered}$$

Therefore, the resistance of a $\text{'}\text{'}100-W\text{'}\text{'}$fluorescent bulb is$626\mathrm{Î\copyright }$