Question

Two electric bulbs marked $40 \,W$, $220 \,V$ and $60 \,W$, $220\, V$ when connected in series, across same voltage supply of $220\, V$, 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,
$R_{S}=R_{B_1}+R_{B_2}=\frac{\left(220\right)^{2}}{40}+\frac{\left(220\right)^{2}}{60}$
$=\left(200\right)^{2} \left[\frac{1}{40}+\frac{1}{60}\right]$
$=\left(220\right)^{2} \left[\frac{60+40}{60\times40}\right]$
$=\left(220\right)^{2}\left(\frac{100}{2400}\right)$
$=\frac{\left(220\right)^{2}}{24}$
$\therefore P_{1}=\frac{V_{s}^{2}}{R_{S}}=\left(200\right)^{2}\times\frac{24}{\left(200\right)^{2}}$
$24\,W $
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 $R_{p}=\frac{\left(200\right)^{2}}{100}$
$\therefore P_{2}=\frac{V_{S}^{2}}{R_{P}}=\left(220\right)^{2}\times\frac{100}{\left(220\right)^{2}}$
$100\,W$
$\therefore \frac{P_{1}}{P_{2}}=\frac{24\,W}{100\,W}$
$=0.24$