A charge of #16 C# passes through a circuit every #12 s#. If the circuit can generate #12 W# of power, what is the circuit's resistance?

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Answer 1 · 2016-04-17T11:31:43

#6.75Omega#

Current is how much charge passes through a point per unit time, or

#I = Q/t#.

In this case, the charge #Q=16C# and the time is #t=12s#, so

#I=16/12=4/3=1.33A#.

Then you can solve for resistance by

#R=P/I^2#.

Power is given in the question as #12W# and #I# we know is #4/3#, so

#R=12/(16/9)=108/16=6.75Omega#