Question

A light bulb is rated $150W$ for $220V$ AC supply of $60Hz$. Calculate
(i) the resistance of the bulb
(ii) the rms current through the bulb
OR
An alternating voltage given by $V=70\sin 100\pi t$ is connected across a pure resistor of $25\Omega$. Find
(i) the frequency of the source
(ii) the rms current through the resistor

Answer 1

(i) Given - $P=150W,V_{rms}=220V,f=60Hz$

by $P=V_{rms}^2/R$

or $R=V_{rms}^2/P={220}^2/150=322.67\Omega$

(ii) now $I_{rms}=V_{rms}/R=220/322.67=0.68A$

OR

(i) If $V=70\sin 100\pi t$

comparing it with $V=V_0\sin \omega t$

$\omega =100\pi$

or $2\pi f=100\pi$

or $f=100/2=50Hz$

(ii) $I_0=V_0/R=70/25=2.8A$

therefore $I_{rms}=I_0/\sqrt{2}=2.8/\sqrt{2}=1.98A$