Question

Two bulbs $A$ and $B$ rated $40W$, $220V$ and $60W$, $220V$, respectively, are connected in series across a $220V$ supply. As a result:

• A. $A$ and $B$ will burn with the same brightness
• B. $A$ burns more brightly
• C. $B$ burns more brightly
• D. Both will be dim

Answer 1

The correct option is B $A$ burns more brightly

$P=\frac{V^2}{R}\Rightarrow V_A=\frac{V^2}{P}=\frac{220\times 220}{40}=1210\Omega$
$R_B=\frac{220\times 220}{60}=806.67\Omega$
In this case, the applied voltage across the series combination itself is $220V$, the rated voltage of both the bulbs.
As a result, potential difference across the bulbs will not exceed the rated value.
At the same time, current through each bulb will be the same.
Hence, the bulb with greater resistance will glow brighter $($since $I^{2}R$ will be greater$)$.
Resistance of bulb $A>$ resistance of bulb $B$.
Therefore, $A$ burns brighter than $B$.