Question

(a) For a given a.c.,$i=i_{m}sin\omega t$, 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 100 W for a 220 V a.c. supply
Calculate the resistance of the bulb

Answer 1

(a) : Current is given as $i=i_{m}sin\left(wt\right)$

Average power dissipated over a complete cycle $<P>=\frac{{\int }_0^T{i}^{2}Rdt}{T}$

$\therefore$ $<P>=\frac{{\int }_0^T{i}_m^{2}si{n}^2\left(wt\right)Rdt}{T}$

OR $<P>=\frac{i_m^{2}R}{T}{\int }_0^{T}si{n}^2\left(wt\right)dt=\frac{i_m^{2}R}{T}{\int }_0^T\frac{1-cos\left(2wt\right)}{2}dt$

OR $<P>=\frac{i_m^{2}R}{2T}\times [t-\frac{sin\left(2wt\right)}{2}]{|}_0^T$

OR $<P>=\frac{i_m^{2}R}{2T}\times [T-0]$ $⟹<P>=\frac{i_m^{2}R}{2}$

(b) : Given $P=100$ W $V=220$ volts

$\therefore$ Resistance of the bulb $R=\frac{(220{)}^2}{100}=484$ $\Omega$