Hi again - partially stuck with a question and I'm not sure where to go. Hoping someone would be able to give me a hint :) A 75 W non-inductive light bulb is designed to run from an ac supply of 120 V rms to 50 Hz. If the only supply available is 240 V rms show that the bulb can be run at the correct power by placing a capacitor C in series with it. Show that the capacitor draws 75 W from the supply and calculate it's capacitance. So I've said that in order for the bulb to still run at the same power, it will need the same charge to flow across it (I'm not sure if this is the right argument though?) Therefore, the power the capacitor draws will be 0.5*Q*V, where V=240 V and Q=75/120 & this gives 75 W. For the capacitance, I said that 75 W will be equal to V^2/2Z, where Z=1/(omega*C); the impedance of the capacitor. This gives me 8.3 micro Farads and the answer is 9.6 micro Farads. What have I done wrong?