A charge of #24 C# passes through a circuit every #9 s#. If the circuit can generate #32 W# of power, what is the circuit's resistance?

Question

1083 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-07T17:22:23

#4.5Omega#

Electric current #I# is the rate of flow of charge:

#I=Q/t=24/9=2.66"A"#

Power #P# is given by:

#P=VxxI=32"W"#

So the voltage drop is given by:

#V=32/I=32/2.66=12"V"#

Using Ohm's Law:

#R=V/I=12/2.66=4.5Omega#