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A charge of #35 C# passes through a circuit every #5 s#. If the circuit can generate #14 W# of power, what is the circuit's resistance?

Answer 1

Emily Abbate

#R=2/7ohm#

#I=(Delta Q)/(Delta t)# #I=35/5=7A# #P=I^2*R# #14=7^2*R# #R=14/(7*7)# #R=2/7ohm#

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Answer 2

Elena Albarran

To find the resistance (R) in the circuit, you can use Ohm's Law:

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

Given that power (P) is 14 W and the charge passing through the circuit every 5 seconds (Q) is 35 C, you can use the formula:

[ P = \frac{Q^2}{R \cdot t} ]

Rearrange the formula to solve for resistance (R):

[ R = \frac{Q^2}{P \cdot t} ]

Substitute the values:

[ R = \frac{(35 , \text{C})^2}{14 , \text{W} \cdot 5 , \text{s}} ]

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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.