Power = 120 W
Voltage = 240 V

120 = I x 240

Divide both sides by 240

I = $\frac{120}{240}$ = 0.5 A

$\frac{1}{{\mathrm{R}}_{\mathrm{total}}}$ = $\frac{1}{5}$ + $\frac{1}{5}$

$\frac{1}{{\mathrm{R}}_{\mathrm{total}}}$ = $\frac{\mathrm{1\; x\; 1\; +\; 1\; x\; 1}}{5}$

$\frac{1}{{\mathrm{R}}_{\mathrm{total}}}$ = $\frac{\mathrm{1\; +\; 1}}{5}$

$\frac{1}{{\mathrm{R}}_{\mathrm{total}}}$ = $\frac{2}{5}$

Cross multiply

R_total x 2 = 1 x 5

Divide both sides by 2

R_total = $\frac{5}{2}$ = 2.5 Ω

Power = 60 W
Voltage = 240 V

Current = $\frac{60}{240}$ = 0.25 A

When R_total is calculated, the result is less than any one of the resistors.

Voltage = 24 V
Resistance = 15 Ω

Current(I) = $\frac{24}{15}$ = 1.6 A

Power which is measured in Watts is work done/energy (J) over time (s)