Question

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 Ω and a power of 30.0 W?

Answer 1

We know that the power is given by the equation:

`P=I^2R`

Plugging the values given for the power and the resistance we have that:

`\begin{gathered} 30=25I^2 \\ I^2=(30)/(25) \\ I=\sqrt[]{(30)/(25)} \\ I=1.09 \end{gathered}`

Therefore the current is 1.09 A.