A charge of #5 C# passes through a circuit every #5 s#. If the circuit can generate #15 W# of power, what is the circuit's resistance?

Answer 1

To find the circuit's resistance, we first calculate the current flowing through the circuit using the formula for current, (I = \frac{Q}{t}), where (Q) is the charge (in coulombs) and (t) is the time (in seconds).

Given:

(Q = 5) C
(t = 5) s

So, (I = \frac{5 , \text{C}}{5 , \text{s}} = 1 , \text{A}).

Next, we use the power formula related to current and resistance, (P = I^2R), where (P) is the power (in watts), (I) is the current (in amperes), and (R) is the resistance (in ohms). We can rearrange this formula to solve for (R):

(R = \frac{P}{I^2}).

Given that the power (P = 15) W and we've calculated (I = 1) A,

(R = \frac{15 , \text{W}}{(1 , \text{A})^2} = 15 , \Omega).

Therefore, the circuit's resistance is (15 , \Omega).

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Answer 2

Nathan Ashcroft

#R = 15 Omega#

#(5 C)/(5 s) = 1# ampere

#P = I²R#

#15 W = (1 A)#R

#R = 15 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.