Answer
(a) $192\Omega$
(b) $0.88A$
(c) $150W$
Work Step by Step
(a) We can find the resistance of the light bulb as follows:
$R=\frac{V_{rms}^2}{P_{avg}}$
We plug in the known values to obtain:
$R=\frac{(120)^2}{75}$
$R=192\Omega$
(b) We can find the maximum current in the bulb as follows:
$I_{rms}=\frac{V_{rms}}{R}$
We plug in the known values to obtain:
$I_{rms}=\frac{120}{192}=0.625A$
Now $I_{max}=\sqrt{2}I_{rms}$
We plug in the known values to obtain:
$I_{max}=\sqrt{2}(0.625)=0.88A$
(c) We can find the maximum power as
$P_{max}=I_{max}^2R$
We plug in the known values to obtain:
$P_{max}=(0.88)^2(192)$
$P_{max}=150W$