Variable resistor in circuit

• The equivalent resistance of the 2 bulbs and resistor in parallel is $\dfrac 1{\frac 14 + \frac 16 + \frac 14} = \dfrac 32 \ \Omega$ .
• The equivalent resistance of the entire circuit at time $t$ , $R(t) = \dfrac 32 + 2 + t = \dfrac 72 + t \ \Omega$ .
• The instantaneous voltage across a bulb $V_b(t) = \dfrac {\frac 32}{\frac 72+t} \times 24 = \dfrac {72}{7+2t} \ V$ .
• Then, the power generated by a bulb when $t=10\text{ s}$ :

$\begin{aligned} P(t) & = \frac {\left(V_b(t)\right)^2}{R_4} & \small \color{#3D99F6} \text{where }R_b \text{ is the resistance of a bulb.} \\ & = \frac {72^2}{4(7+2t)^2} \\ P(10) & = \frac {72^2}{4(27^2)} \\ & = \frac {4^2}{3^2} \text{ W} \end{aligned}$

Therefore, $a+b=4+3 = \boxed 7$ .

A good follow-up would be to ask for the total energy dissipated in one bulb from $t= 0$ to $t = 10$