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:

$\begin{gathered} 30=25I^2 \\ I^2=\frac{30}{25} \\ I=\sqrt[]{\frac{30}{25}} \\ I=1.09 \end{gathered}$

Therefore the current is 1.09 A.