- #1
lollol
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Can someone give me a reply Intro Physics style? :D
P = IV is known as the "power of the circuit" right?
But then, we have questions asking about the heat dissipated by a resistor. We can use P = IV, P=I^2 R , or P=V^2/R to answer such questions
Since the voltage drops all add up to the original voltage.. won't adding IV + IV for each resistor yield the original Power... implying that ALL energy is dissipated as heat?
I guess I don't understand the difference between the generic "Power" of the circuit and the power/head dissipated by resistors :(
P = IV is known as the "power of the circuit" right?
But then, we have questions asking about the heat dissipated by a resistor. We can use P = IV, P=I^2 R , or P=V^2/R to answer such questions
Since the voltage drops all add up to the original voltage.. won't adding IV + IV for each resistor yield the original Power... implying that ALL energy is dissipated as heat?
I guess I don't understand the difference between the generic "Power" of the circuit and the power/head dissipated by resistors :(