A charge of #35 C# passes through a circuit every #5 s#. If the circuit can generate #21 W# of power, what is the circuit's resistance?

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Answer 1 · 2018-05-03T12:53:28

#3/7~~0.43# ohms

We first find the current produced in five seconds. Current is given by the equation,

#I=Q/t#

#Q# is the charge in coulombs

#t# is the time in seconds

So, we get:

#I=(35 \ "C")/(5 \ "s")#

#=7 \ "A"#

Now, power is related by the equation,

#P=I^2R#, since #P=IV# and #V=IR# (Ohm's law)

#I# is the current in amperes

#R# is the resistance in ohms

#P# is the power in watts

Rearranging for resistance, we get:

#R=P/I^2#

Plugging in our given values, we get:

#R=(21 \ "W")/((7 \ "A")^2)#

#=(21 \ "W")/(49 \ "A"^2)#

#=3/7 \ Omega#