Question

A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.

Answer 1

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, E_rms = 220 V
P = V²R,
where R = resistance of the bulb
$$\begin{gathered} \therefore R\mathit{=}\frac{V^{\mathit{2}}}{P}=\frac{220\times 220}{60} \\ =806.67 \end{gathered}$$
Peak value of voltage $\left(E_0\right)$ is given by,
$E_0=E_{\mathrm{rms}}\sqrt{2}$
=$220\times \sqrt{2}$
= 311.08
Now, maximum current through the filament $\left(i_0\right)$ is,
$i_0=\frac{E_0}{R}$
$\Rightarrow i_0=\frac{311.08}{806.67}=0.39\mathrm{A}$