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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?

Answer 1

William Audy

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

The charge that flows through a point once every second is equal to the current, or I.

I= 6 C/s = 6 Amperes / 6 S = 36C / 6 S.

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#

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Answer 2

Layla Ahmed

The circuit's resistance is 1 ohm.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.