My HW problem states: a) Find the internal resistance of a 60W light bulb. Note: the power rating of a light bulb is determined by measuring the real power dissipated when the standard, 120V/60Hz AC outlet voltage is applied to the bulb. b) The light bulb is connected in series with a 20Ω resistor and the standard AC outlet voltage is applied to the series in combination. What is the real power dissipated on the light bulb? c) what is the real power dissipated on the resistor? So, This is what I got, but im not sure what the difference is between regular power and real power: a) P = V^2 / R R = 120^2 / 60W = 240Ω b) I first found the current, I I = V / R (where R = Req = 260Ω) I = 120V / 260Ω = .462A Then, real power = (I^2)(R) so P = (.462A^2)(240Ω) = 51.23W c) P = (.462A^2)(20Ω) = 4.27W are these the right answers?