Answer
See answers.
Work Step by Step
We use 2 significant figures.
a. Current is power over voltage, from equation 18–5.
$$I_A=P_A/V_A=40W/120V=0.33A$$
$$I_B=P_B/V_B=40W/12V=3.3A$$
b. Resistance is calculated from equation 18–6b.
$$R_A=V_A^2/P_A=(120V)^2/40W=360 \Omega$$
$$R_B=V_B^2/P_B=(12V)^2/40W=3.6 \Omega$$
c. The charge is the current multiplied by the time.
$$Q_A=I_A t=(0.33A)(3600s)=1200C$$
$$Q_B=I_B t=(3.3A)(3600s)=1.2\times10^4 C$$
d. The energy is the power multiplied by the time. The power is 40W for both bulbs.
$$E=Pt=(40W)(3600s)=1.4\times10^5 J$$
e. Bulb B requires larger-diameter wires because it draws more current.