A charge of #30 C# passes through a circuit every #5 s#. If the circuit can generate #12 W# of power, what is the circuit's resistance?

Answer 1

#R=1/3 ohm#

#"The electric current is determined by:"# #I=(Delta Q)/(Delta t)# #"I:electric current"# #"Q:charge"# #"t:time"# #I=30/5# #I=6A# #"P:Power ;R: resistance of circuit"# #P=I^2R# #12=36*R# #R=1/3 ohm#

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Answer 2

Adam Arnold

Use the formula: ( P = \frac{V^2}{R} ) where ( P ) is power, ( V ) is voltage, and ( R ) is resistance. Rearrange to find ( R ): ( R = \frac{V^2}{P} ). Given ( P = 12 , \text{W} ) and ( Q = 30 , \text{C} ) passing every ( 5 , \text{s} ), ( V = \frac{Q}{t} ). Substitute values into ( R ) formula to get resistance. ( R \approx 2 , \Omega ).

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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.