NEET-XII-Physics
39: Alternating Current
- #3A 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.Ans : Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, Erms = 220 V
P = V2R,
where R = resistance of the bulb
`` \therefore R\mathit{=}\frac{{V}^{\mathit{2}}}{P}=\frac{220\times 220}{60}``
`` =806.67``
`` ``
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\,}``
Page No 330: