Question

(a) For a given AC, $i=i_{m}sin\omega t$, show that the average power dissipated in a resistor $R$ over a complete cycle is $\frac{1}{2}{i}_m^{2}R$.
(b) A light bulb is rated at $100W$ for a $220V$ AC supply. Calculate the resistance of the bulb.

Answer 1

(a) The average power dissipated
$P=|i^{2}R|=|i_m^{2}Rsi{n}^2\omega t|$
$=i_m^{2}R|si{n}^2\omega t|$
$si{n}^2\omega t=\frac{1}{2}(1-cos2\omega t)$
$\Rightarrow |si{n}^2\omega t|=\frac{1}{2}(1-|cos2\omega t|)=\frac{1}{2}[\because (|cos2\omega t|)=0]$
$\therefore \overline{P}=\frac{1}{2}{i}_m^{2}R$
(b) Power of the bulb, $P=100W$ and voltage, $V=220V$
The resistance of the bulb is given as
$R=\frac{V^2}{P}$
$=\frac{(220{)}^2}{100}$
$=484\Omega$