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OpenStudy (anonymous):
what is the power dissapation for a series circuit that has 10V ,1.0 Kohms ,6.8 Kohms
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OpenStudy (anonymous):
\[Power=V*I..........V=I*R...........Power=\frac{V^2}{R}\]
OpenStudy (anonymous):
\[Power=\frac{10^2}{1000 ohms}=0.1 watts.... and.... Power=\frac{10^2}{6800}=1.47x10^{-2} watts\]
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