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The power in a lightbulb is given by the equation [tex]P = I^2 R[/tex], where [tex]I[/tex] is the current flowing through the lightbulb and [tex]R[/tex] is the resistance of the lightbulb. What is the current in a circuit that has a resistance of [tex]30.0 \Omega[/tex] and a power of 2.00 W?

A. 15.0 A
B. 0.067 A
C. 0.258 A
D. 3.87 A

Sagot :

GinnyAnsweridn
To determine the current flowing through the circuit, we start with the power equation for an electrical circuit:

[tex]\[ P = I^2 R \][/tex]

where:
- [tex]\( P \)[/tex] is the power in watts (W)
- [tex]\( I \)[/tex] is the current in amperes (A)
- [tex]\( R \)[/tex] is the resistance in ohms (Ω)

We need to solve for the current [tex]\( I \)[/tex]. Rearranging the power equation to solve for [tex]\( I \)[/tex]:

[tex]\[ I^2 = \frac{P}{R} \][/tex]
[tex]\[ I = \sqrt{\frac{P}{R}} \][/tex]

Substituting the given values into the equation:
- [tex]\( P = 2.00 \, \text{W} \)[/tex]
- [tex]\( R = 30.0 \, \Omega \)[/tex]

[tex]\[ I = \sqrt{\frac{2.00}{30.0}} \][/tex]

Now, let's calculate it step-by-step.

1. First, perform the division inside the square root:

[tex]\[ \frac{2.00}{30.0} = 0.0667 \][/tex]

2. Next, take the square root of the result:

[tex]\[ I = \sqrt{0.0667} \approx 0.258 \, \text{A} \][/tex]

Thus, the current in the circuit is approximately [tex]\( 0.258 \)[/tex] A.

So, the correct answer is:
C. 0.258 A
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