Question

Two bulbs, one of $200\text{Volts},60\text{Watts}$ and the other of $200\text{Volts},100\text{Watts}$ are connected in series to a $200\text{Volt}$ supply. The power consumed will be

• A. $37.5\text{Watt}$
• B. $160\text{Watt}$
• C. $62.5\text{Watt}$
• D. $110\text{Watt}$

Answer 1

The correct option is A $37.5\text{Watt}$

Let us find the resistances using the given ratings and formula will be

$P_{rated}=\frac{{\left(V_{rated}\right)}^2}{R}\Rightarrow R=\frac{{\left(V_{rated}\right)}^2}{P_{rated}}$

$\therefore R_1=\frac{{\left(200\right)}^2}{60}\Omega ;R_2=\frac{{\left(200\right)}^2}{100}\Omega$


Since, both are in series, so effective resistance will be

$R_{eq}=R_1+R_2$

So, total power consumed by circuit will be

$P_{net}=\frac{V^2}{R_{eq}}=\frac{{\left(200\right)}^2}{R_1+R_2}$

$P_{net}=\frac{{\left(200\right)}^2}{(200{)}^2[\frac{1}{60}+\frac{1}{100}]}=37.5\text{W}$

Hence, option $\left(a\right)$ is correct.