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.
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
discusses the physics principles used in this problem. Three resistors, 2.72, 3.03, and 4.32Ω, are connected...
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