SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Given, device rating is 120 V, 100 W
So, from (1), resistance of device
${\mathrm{R}}_{\mathrm{d}}=\frac{{\left(120\right)}^2}{100}=144\mathrm{\Omega }$
Now let's add external resistor ${\mathrm{R}}_{\mathrm{ex}}$ in series with device resistance ${\mathrm{R}}_{\mathrm{d}}$.Line voltage of device reduces from 120 to $72\mathrm{V}$, so the remaining voltage $\left(120-72=48\mathrm{V}\right)$ should drop across ${\mathrm{R}}_{\mathrm{ex}}$.
Current through circuit,
$\mathrm{I}=\frac{72}{144}=\frac{48}{{\mathrm{R}}_{\mathrm{ex}}}$
${\mathrm{R}}_{\mathrm{ex}}=96\mathrm{\Omega }$