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A charge of #54 C# passes through a circuit every #6 s#. If the circuit can generate #162 W# of power, what is the circuit's resistance?

Answer 1

Lily Adams

#R=2" "Omega#

#I=(Delta Q)/(Delta t)" Current of circuit"#

#Delta Q=54C" ; "Delta t=6 s#

#I=54/6=9 " "A#

#P:"Power"#

#R:"Resistance"#

#P=I^2*R#

#162=9^2*R#

#R=162/81#

#R=2" "Omega#

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Answer 2

Leo Abeyta

I found #2Omega#

You know that current #I# is the charge passing a section of conductor each second. In your case you have: #I=Q/t=54/6=9A# Power is the rate of transformation (work) energy with time: #P=E/t=V*I# Using Ohm's Law we get: #P=I^2R# rearranging: #R=P/I^2=162/(9^2)=2Omega#

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Answer 3

Axel Acevedo

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.