- #1

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## Homework Statement

Just problem 19C.

## Homework Equations

P=IV=Ie

^{iwt}*Ve

^{iwt}. T

## The Attempt at a Solution

P = IVe

^{2iwt}=IVcos(2wt). What did I do wrong?

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- Thread starter gimak
- Start date

- #1

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Just problem 19C.

P=IV=Ie

P = IVe

- #2

mjc123

Science Advisor

Homework Helper

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It asks for the time average. Integrate it over a cycle, i.e. t = 0 to 2π/ω.

- #3

ehild

Homework Helper

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## Homework Statement

View attachment 205836

Just problem 19C.

=

## Homework Equations

P=IV=Ie^{iwt}*Ve^{iwt}. T

## The Attempt at a Solution

P = IVe^{2iwt}=IVcos(2wt). What did I do wrong?

The instantaneous power is the product of the real voltage and the real current. The complex form of power and voltage can be used for linear operations only (addition, multiplication with constant). In general, the current and voltage need not be in phase.

The complex voltage

The instantaneous power is the product or the real part of the complex current (I(t)=I

- #4

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The complex form of power and voltage can be used for linear operations only (addition, multiplication with constant).

So this means that we can't use their complex form because whatever operators we use on them are only linear? Also, is another way to understand this is that since power & voltage are real, that means we must take their real part when doing operations with them?

- #5

ehild

Homework Helper

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You can do linear operations with the complex voltages and currants, but in any other case, the real quantities should be used.So this means that we can't use their complex form because whatever operators we use on them are only linear? Also, is another way to understand this is that since power & voltage are real, that means we must take their real part when doing operations with them?

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