A charge of #45 C# passes through a circuit every #6 s#. If the circuit can generate #98 W# of power, what is the circuit's resistance?

1 Answer
Scott S.
Sep 15, 2016

The resistance is 1.742 #Omega#

Explanation:

The current in the circuit is the amount of charge that passes by a point per second. It is found from the equation #I = (Delta Q)/(Delta t)= (45 C)/(6 s) = 7.5 A#.

The power in a circuit is found from the equation #P = IV = I^2R#. This means that the resistance is #R = (P)/(I^2) = (98 W)/(7.5 A)^2 = 1.742 Omega#

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