Question

question 8 of 25
the power in a lightbulb is given by the equation ( p = i^{2}r ), where ( i ) is the current
flowing through the lightbulb and ( r ) is the resistance of the lightbulb. what is
the current in a circuit that has a resistance of ( 25.0 omega ) and a power of ( 30.0 w )?
a ( 0.910 a )
b. ( 1.20 a )
c. ( 1.09 a )
d. ( 0.830 a )

Explanation:

Step1: Rearrange the formula

Given \(P = I^{2}R\), we can solve for \(I\) by \(I=\sqrt{\frac{P}{R}}\).

Step2: Substitute values

Substitute \(P = 30.0\space W\) and \(R=25.0\space\Omega\) into the formula: \(I=\sqrt{\frac{30.0}{25.0}}\).

Step3: Calculate

\(\frac{30.0}{25.0}=1.2\), then \(I = \sqrt{1.2}\approx1.09\space A\).

Answer:

C. \(1.09\space A\)