91Ó°ÊÓ

We need to find the order of brightness of the bulbs, from brightest to dimmest.

The energy is dissipated when the current flows through a resistor. The rate of the dissipated energy is the power. It is given as

$P_R=\frac{(∆V_R)}{R}$

The bulbs are identical, so the resistance will be the same. Calculating the voltage for each bulb. The combination $ST$is in parallel with the other branch. So, the voltage across the right branch and the middle branch is the same $∈$.

For the right branch, the voltage is divided by the two resistors $S$and $T$, so the voltage across each one is given as

$∆V_S=∆V_T=\frac{∈}{2}$

As both resistors have the same voltage and the same current, they will have the same brightness$S=T$.

For the middle branch, the total voltage is $∈$, but the combination $QR$has resistance less than bulb $P$.So, the bulb $P$has larger voltage than the combination $QR$.Hence, the voltage of $P$is $\frac{2}{3}∈$ while the combination $QR$has voltage $\frac{1}{3}∈$because the summation of the values is $\frac{2}{3}∈+\frac{1}{3}∈=∈$.

The combination $QR$has voltage $\frac{1}{3}∈$, so each bulb has the same voltage by

$∆V_Q=∆V_R=\frac{1}{6}∈$

From the results, we can order the voltage by

$∆V_P>∆V_S=∆V_T>∆V_Q=∆V_R$

More voltage gives more brightness. So, the correct answer is

$P>S=T>Q=R$