Variable Resistor and Bulb Energy

• 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 energy dissipated in a bulb from $t=0$ to $t=10\text{ s}$ :

$\begin{aligned} E & = \int_0^{10} \frac {\left(V_b (t)\right)^2}{R_b} dt & \small \color{#3D99F6} \text{where }R_b \text{ is the resistance of the bulb.} \\ & = \int_0^{10} \frac {1296}{(7+2t)^2} dt \\ & = \frac {648}{7+2t} \bigg|_{10}^0 \\ & = 648 \left(\frac 17 - \frac 1{27}\right) \\ & \approx \boxed{68.57} \end{aligned}$