A current of 3 A flows through a resistor for 6 minutes. Determine the power dissipated in the resistor if 3V existed across the resistor.
Power
P=i2RP=i^2RP=i2R
R=VI=33=1AR=\frac{V}{I}=\frac{3}{3}=1AR=IV=33=1A
P=32×1=9WP=3^2\times1=9WP=32×1=9W
Work(W)=Pt=9×360=3240JPt=9\times 360=3240JPt=9×360=3240J
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