The stereo uses an energy of [tex]E=1950 J[/tex] in a time [tex]t=1 h=3600 s[/tex], therefore the power of the stereo is given by
[tex]P= \frac{E}{t}= \frac{1950 J}{3600 s}=0.54 W [/tex]
We also know that the power of an electrical device is related to its voltage, V, and its resistance, R, by the following equation
[tex]P= \frac{V^2}{R} [/tex]
therefore, we can rearrange the equation to calculate the resistance of the stereo:
[tex]R= \frac{V^2}{P}= \frac{(4.5 V)^2}{0.54 W}=37.5 \Omega [/tex]
