91Ó°ÊÓ

The brightness of the bulb depends upon the amount of energy per second dissipated as light through the bulb when current pass through it.

The equation for the power of the bulb is,

$P=I^{2}R$

Here, $I$is the current in the circuit and $R$is the resistance of the bulb.

An ordinary light bulb dissipates surprisingly little power, in comparison with bulbs $b$and $c$. Both bulbs b and c have a resistance that doubles that of an individual bulb.

To make sure you are convinced, let us point out that the power dissipated in a resistor when the potential difference is given is,

$P=UI=U\frac{U}{R}=\frac{U^2}{R}$

Therefore, we can conclude that both bulbs $b$and$c$ dissipate the same amount of power. Due to their identical currents and resistances, following can be concluded.
$P=UI=IRI=I^{2}R.$

The rank is,

$a>b=c$

Hence, the rank in order, from brightest to least bright light bulb is$a>b=0.$