Question

Two electric bulbs have ratings respectively of $25W,220V$ and $100W,220V$. If the bulbs are connected in series with a supply of $440V$, which bulb will fuse?

• A. $25W$ bulb
• B. $100W$ bulb
• C. Both of these
• D. None of these

Answer 1

The correct option is A $25W$ bulb

Power is given by $P=\frac{V^2}{R}$, where $P$ is Power, $V$ is voltage across bulb and $R$ is resistance of bulb.

Hence, $R=\frac{V^2}{P}$

Resistance of $25Wbulb=\frac{220\times 220}{25}=1936\Omega$

Using Ohm's law, $V=iR$
Value of current that can pass through it $i=\frac{V}{R}=\frac{220}{1936}=0.11A$

Resistance of $100Wbulb=\frac{220\times 220}{100}=484\Omega$
Value of current that can pass through it $i=\frac{V}{R}=\frac{220}{484}=0.45A$

When connected in series to $440V$ supply, the equivalent resistance is given by,

$R_{equi}=R_1+R_2=1936+484=2420\Omega$

Calculating the value of the current $i=\frac{440}{2420}=0.18A$

Thus $0.18A>0.11A$

The value of current flowing in the final circuit is greater than that allowed by the $25Wbulb$, so it will fuse.