Answer:
The total power is 161 W
Explanation:
Using the equation bellow we can write the total power delivered to this equivalent resistance
P = [tex]\frac{V^{2} }{R1 + R2}[/tex]
According to this equation, when each heater is connected by itself to the battery, we have
[tex]P1 = \frac{V^{2} }{R1}[/tex] or [tex]R1 = \frac{V^{2} }{P1}[/tex]
[tex]P2 = \frac{V^{2} }{R2}[/tex] or [tex]R2 = \frac{V^{2} }{P2}[/tex]
Substituting both equation into the first one
[tex]P = \frac{V^{2} }{\frac{V^{2} }{P1} + \frac{V^{2} }{P2} } = \frac{1}{\frac{1}{P1} + \frac{1}{P2} } = \frac{P1P2}{P1+P2} = \frac{380W*280W }{380W+280W} = 161W[/tex]
