A charge of #5 C# passes through a circuit every #2 s#. If the circuit can generate #15 W# of power, what is the circuit's resistance?

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Answer 1 · 2017-11-06T21:53:38

The resistance is #2.4 Omega#

A rate of charge movement that causes 5C of charge to pass every 2 s is a current of 2.5 A

#I=Q/t = (5 C)/(2 s) = 2.5 C/s or 2.5 A#

Next we use the relation

#P=I^2R# to find the resistance

Write this as

#R=P/I^2 = (15W)/(2.5A)^2 = 2.4 Omega#

Answer 2 · 2017-11-06T21:54:24

#2.4 Omega#

#C = It#
charge = current * time

#I = C/t = 5/2A#
current = charge/time

#P = I^2R#
power = current^2 * resistance

#15W = (5/2)^2 * R#

#15W = 6.25 * R#

#R = 15/6.25 Omega#

#= 60/25 = 12/5#

#=2.4 Omega#