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Q.
Two bulbs of wattage $40 \,W$ and $100 \,W$ each rated at $200 \,V$ are connected in series across a $440 \,V$. What will happen?
J & K CETJ & K CET 2001
Solution:
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, $P =40 \,W$
$ \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 $100 \,W$ 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 $440 \,V$,
then current $I=\frac{440}{R}=\frac{440}{1400}=0.31 \,A$
Since, $I>I_{1}$ but less than hence the bulb of $40 \,W$ will fuse.