A charge of #25 C# passes through a circuit every #4 s#. If the circuit can generate #25 W# of power, what is the circuit's resistance?

Question

1141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-26T09:58:23

#0.64# ohms.

We first find the current passing through, which is given by the equation:

#I=Q/t#

where:

#I# is the current in amperes

#Q# is the charge in coulombs

#t# is the time in seconds

So, the current is:

#I=(25 \ "C")/(4 \ "s")#

#=6.25 \ "A"#

Power is related through current and resistance by the equation:

#P=I^2R#

where:

#P# is the power in watts

#I# is the current in amperes

#R# is the resistance in ohms

Rearranging for resistance, we get:

#R=P/I^2#

Plugging in the values, we get:

#R=(25 \ "W")/((6.25 \ "A")^2)#

#=(25 \ "W")/(39.0625 \ "A"^2)#

#=0.64 \ Omega#