(a) the power dissipated by the resistor
"P=IV=(4\\:\\rm A)\\times (32\\:\\rm V)=128\\:\\rm W."(b) the amount of energy dissipated by the resistor in 20 s
"E=Pt=(128\\:\\rm W)\\times (20\\:\\rm s)=2560\\:\\rm J."(c) the resistance of the resistor
"R=\\frac{V^2}{P}=\\frac{(32\\:\\rm V)^2}{(128\\:\\rm W)}=8\\:\\rm \\Omega."
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Dear visitor, please use panel for submitting new questions
Dear visitor, please use panel for submitting new questions
For the electric circuit above calculate: a) The equivalent resistance R!". b) The electric current through the battery. c) The potential difference across the resistors of 3 kΩ and 2 kΩ.
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