An AC voltage source of variable angular frequency ω and fixed amplitude V₀ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased, what happens to the bulb's brightness and the total impedance of the circuit?

Impedance Z = √(1/(ωC)² + R²)

As ω increases, the term 1/(ωC)² decreases. This means the total impedance Z decreases. Since the voltage V₀ is constant, the current I = V₀/Z will increase as Z decreases. The power dissipated by the bulb is given by P = I²R. Since I increases, the power P, and hence the brightness of the bulb, will increase.

Therefore, when ω is increased, the bulb glows brighter and the total impedance of the circuit decreases.