Question

discusses the physics principles used in this problem. Three resistors, 2.72, 3.03, and 4.32Ω, are connected in series across a 25.0-V battery. Find the power delivered to each resistor.

Answer 1

since resistors are connected in series

Rnet = R1 + R2 + R3

= 2.72+3.03+4.32

=10.07 ohm

from OHm's law

V= IReq

I = V/ R eq = 25/10.07 ohm

=2.48 A

power delivered to each resistor is

P1 = I^2 R1

= (2.48 A)^2 ( 2.72)

=16.76 W

P2 = I^2 R2

= (2.48)^2 ( 3.03)

=18.63 W

P3 = I^2 R3

= (2.48)^2 (4.32)

=26.56 W