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]25.0 \Omega[/tex] and a power of [tex]30.0 W[/tex]?

A. [tex]0.830 A[/tex]

B. [tex]0.910 A[/tex]

C. [tex]1.20 A[/tex]

D. [tex]1.09 A[/tex]



Answer :

To determine the current \( I \) in a circuit with a resistance of \( 25.0 \, \Omega \) and a power of \( 30.0 \, \text{W} \), we start with the given power formula:
[tex]\[ P = I^2 R \][/tex]

In this equation:
- \( P \) is the power in watts (\( 30.0 \, \text{W} \)),
- \( I \) is the current in amperes,
- \( R \) is the resistance in ohms (\( 25.0 \, \Omega \)).

We need to solve for the current \( I \). Rearrange the formula to isolate \( I \):

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

Now, substitute the given values for \( P \) and \( R \):

[tex]\[ I^2 = \frac{30.0}{25.0} \][/tex]

Calculate the right side:

[tex]\[ I^2 = 1.2 \][/tex]

To find \( I \), take the square root of both sides:

[tex]\[ I = \sqrt{1.2} \][/tex]

The calculated value of \( \sqrt{1.2} \) is approximately:
[tex]\[ I \approx 1.0954451150103321 \, \text{A} \][/tex]

So, the current in the circuit is approximately \( 1.09 \, \text{A} \). Thus, the correct answer is:

D. [tex]\( 1.09 \, \text{A} \)[/tex]