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A charge of #35# #C# passes through a circuit every #5# #s#. If the circuit can generate #7# #W# of power, what is the circuit's resistance?

Answer 1

Nora Arzate

Convert the flow of charge into current, then use and rearrange #P=I^2R# to find that the resistance is #1/7# #Omega#.

The definition of current is the rate at which charge flows:

#I=Q/t=35/5=7# #A#

An expression for a circuit's power is as follows:

#P=I^2R#

(Combining the equation #P=VI# with Ohm's Law, #V=IR# will yield this.)

Organizing:

#R = P/I^2=7/7^2=1/7# #Omega#

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Answer 2

Adrian Ayala

Use the formula ( P = \frac{V^2}{R} ), where ( P ) is power (7 W), ( V ) is voltage, and ( R ) is resistance. Voltage (( V )) is calculated using ( V = Q / t ), where ( Q ) is charge (35 C) and ( t ) is time (5 s). Substitute these into the power formula to find resistance (( R )).

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

[ V = \frac{35 , \text{C}}{5 , \text{s}} = 7 , \text{V} ]

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

[ R = \frac{7^2}{7} = 7 , \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.