A charge of #24 C# passes through a circuit every #6 s#. If the circuit can generate #8 W# of power, what is the circuit's resistance?

2 Answers
Feb 19, 2016

The resistance in the circuit is #0.5# #Omega#

Explanation:

Data:

Charge#=Q=2C#
Time#=t=6s#
Power#=P=8W#
Resistance#=R=??#

We know that:

#P=I^2R#

Where #I# is the current.

Also we know that: #I=Q/t=24/6=4# #A#

#P=I^2R implies 8=4^2*R#

Rearranging:

#R=8/16=0.5# #Omega#

Hence, the resistance in the circuit is #0.5# #Omega#.

The circuit's resistance is #0.5# #Omega#

Explanation:

First, let's think about what we are trying to calculate. We know the power generated from the circuit which is given by Joule's law:

#P=I*V = 8# #W#

We have also been given the rate of charge flow, which is a current:

#I = Q/t = (24C)/(6s) = 4# #A#

We can substitute this current into the first equation and solve for the voltage:

#V = (8W)/(4A) = 2# #V#

Now we have a current and a voltage and want to find a resistance. We know Ohm's Law relates all three of these:

#V=I*R#

Rearranging to find the resistance:

#R = V/I = (2V)/(4A) = 0.5# #Omega#