Question

Two bulbs of wattage $40W$ and $100W$ each rated at $200V$ are connected in series across a $440V$. What will happen?

• A. 40 W bulb will fuse
• B. 100 W bulb will fuse
• C. Both bulbs will fuse
• D. Nothing will happen

Answer 1

Answer: (A)

From Joules law, power consumed by a bulb of resistance $R$ is
$P=\frac{V^2}{R}$
where $V$ is potential difference.
Given, $V=200$ volt, $P=40W$
$\therefore R_1=\frac{V^2}{P}=\frac{(200{)}^2}{40}=1000\Omega$
From Ohms law, $V=I_{1}R$
$\therefore I_1=\frac{V}{R}=\frac{200}{1000}=\frac{1}{5}A$
Similarly, resistance of $100W$ bulb,
$R_2=\frac{(200{)}^2}{100}\Omega =400\Omega$
$\therefore I_2=\frac{1}{2}A$
when bulbs are connected in series, effective resistance
$R=R_1+R_2=1000\Omega +400\Omega =1400\Omega$
When supply voltage is $440V$,
then current $I=\frac{440}{R}=\frac{440}{1400}=0.31A$
Since, $I>I_1$ but less than hence the bulb of $40W$ will fuse.