# A charge of #16 C# passes through a circuit every #12 s#. If the circuit can generate #1 W# of power, what is the circuit's resistance?

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To find the resistance of the circuit, you can use the formula:

[ P = \frac{V^2}{R} ]

Given that power ( P = 1 , W ) and charge ( Q = 16 , C ) passes through the circuit every ( t = 12 , s ), you can calculate voltage using the formula:

[ V = \frac{Q}{t} ]

Then, rearrange the power formula to solve for resistance:

[ R = \frac{V^2}{P} ]

[ R = \frac{\left(\frac{Q}{t}\right)^2}{P} ]

[ R = \frac{\left(\frac{16}{12}\right)^2}{1} ]

[ R = \frac{\left(\frac{4}{3}\right)^2}{1} ]

[ R = \frac{\frac{16}{9}}{1} ]

[ R = \frac{16}{9} ]

[ R = 1.777 , \Omega ]

So, the circuit's resistance is approximately ( 1.777 , \Omega ).

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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.

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.

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.

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