Three resistors, 1.17, 2.84, and 3.65Ω, are connected in series across a 20.3-V battery. Find the power delivered to each resistor.

Answer 1

R=R₁+R₂+R₃=1.17+2.84+3.65 =7.66 Ω
I=U/R = 20.3/7.66=2.65 A
P₁ =I²R₁=2.65²•1.17=
P₂ =I²R₂=2.65²•2.84=
P₃=I²R₃=2.65²•3.65=