Question

Two electric bulbs marked $40W$, $220V$ and $60W$, $220V$ when connected in series, across same voltage supply of $220V$, the effective power is $P_1$ and when connected in parallel, the effective power is $P_2$. Then $\frac{p_1}{P_2}$ is

• A. $0.5$
• B. $0.48$
• C. $24$
• D. $0.16$
• E. $4.1$

Answer 1

Answer: (C)

Resistance of a bulb $=\frac{\text{(Rated voltage}{)}^2}{\text{(Rated power)}}$
$R_{B_1}=\frac{{\left(220\right)}^2}{40}\Omega$ and $R_{B_2}=\frac{{\left(220\right)}^2}{60}\Omega$
When the bulbs are connected in series,
$R_S=R_{B_1}+R_{B_2}=\frac{{\left(220\right)}^2}{40}+\frac{{\left(220\right)}^2}{60}$
$={\left(200\right)}^2\left[\frac{1}{40}+\frac{1}{60}\right]$
$={\left(220\right)}^2\left[\frac{60+40}{60\times 40}\right]$
$={\left(220\right)}^2\left(\frac{100}{2400}\right)$
$=\frac{{\left(220\right)}^2}{24}$
$\therefore P_1=\frac{V_s^2}{R_S}={\left(200\right)}^2\times \frac{24}{{\left(200\right)}^2}$
$24W$
When the bulbs are connected in parallel
$\frac{1}{R_p}=\frac{1}{R_{B_1}}+\frac{1}{R_{B_2}}$
or $\frac{1}{R_P}=\frac{40}{{\left(220\right)}^2}+\frac{60}{{\left(220\right)}^2}$
$\frac{1}{RP}=\frac{100}{{\left(200\right)}^2}$
or $R_p=\frac{{\left(200\right)}^2}{100}$
$\therefore P_2=\frac{V_S^2}{R_P}={\left(220\right)}^2\times \frac{100}{{\left(220\right)}^2}$
$100W$
$\therefore \frac{P_1}{P_2}=\frac{24W}{100W}$
$=0.24$