A charge of #36 C# passes through a circuit every #6 s#. If the circuit can generate #6 W# of power, what is the circuit's resistance?

1 Answer
Feb 18, 2016

#R = (6W)/(6A)^2= (6W)/(36A) = 1/6text( Ohms) ~= 0.17 Omega#
The circuit's resistance is around 0.17 Ohms.

Explanation:

The current, I, is equal to the charge passing through a point, every second.

I = dq / ds = 36C / 6 s = 6 C/s = 6 Amperes.

Now use, Power = P = #I^2 R#, and rearrange to make the resistance, R, the subject:

#R = P/I^2#
Enter known values for P and I.

#R = (6W)/(6A)^2= (6W)/(36A) = 1/6text( Ohms) ~= 0.17 Omega#