Power calculations of AC signal

[PhysicsTest]

Homework Statement

The voltage across a 10 ohm resistor is v(t) = 170 sin (377t) V. Determine (a) an expression for instantaneous power absorbed by the resistor, (b) the peak power, and (c) the average power.

Relevant Equations

Instantaneous power $p(t) = v(t) i(t)$
Average power $p(t)_{avg} = \frac{\int_{t_0}^{t_0+T} p(t)dt} T$

The below are the formulas i have seen from the book, but no particular formula for the peak power
Instantaneous power $p(t) = v(t) i(t) W$ ->eq1
Average power $p(t)_{avg} = \frac{\int_{t_0}^{t_0+T} p(t)dt} T W$ ->eq2
a. Instantaneous power
$v(t) =A\sin(\omega t)$ ->eq3
$i(t) = \frac{v(t)} R$ -> eq4 R is resistance = 10 Ohm.
Hence the instantaneous power is using eq1
$p(t) = \frac{A^2\sin^2(\omega t)} R = \frac{A^2(1+\cos(2\omega t))} {2R}$W ->eq5 is the answer for part a

b. Peak power
I assume it is not dependent on the variable "t" and it is a constant value. To obtain peak power substitute
$\cos(2\omega t) = 1$. Then the peak power is
$p_{peak} = \frac{A^2} R$ -> eq6, but one doubt here is if it is a very complex expression how do i calculate the peak power?

c. Average power
Using eq2
$p_{avg} = \frac{\int_0^{2\pi} A^2(1+\cos(2\omega t))dt} {2R\pi}$
$p_{avg} = \frac{A^2*2\pi + \frac{[\sin(2\omega t)]_0^{2\pi}} {2\omega}} {2R\pi}$ The sine term will be 0. Hence the expression reduces to
$p_{avg} = \frac{A^2*2\pi} {2R\pi} = \frac{A^2} R$ ->eq7

But peak power and average power are same, I think they should not be same, what is the mistake?

 

[Steve4Physics]

a. Instantaneous power
$v(t) =A\sin(\omega t)$ ->eq3
$i(t) = \frac{v(t)} R$ -> eq4 R is resistance = 10 Ohm.
Hence the instantaneous power is using eq1
$p(t) = \frac{A^2\sin^2(\omega t)} R = \frac{A^2(1+\cos(2\omega t))} {2R}$W ->eq5 is the answer for part a

b. Peak power
I assume it is not dependent on the variable "t" and it is a constant value. To obtain peak power substitute
$\cos(2\omega t) = 1$. Then the peak power is
$p_{peak} = \frac{A^2} R$ -> eq6, but one doubt here is if it is a very complex expression how do i calculate the peak power?

c. Average power
Using eq2
$p_{avg} = \frac{\int_0^{2\pi} A^2(1+\cos(2\omega t))dt} {2R\pi}$
$p_{avg} = \frac{A^2*2\pi + \frac{[\sin(2\omega t)]_0^{2\pi}} {2\omega}} {2R\pi}$ The sine term will be 0. Hence the expression reduces to
$p_{avg} = \frac{A^2*2\pi} {2R\pi} = \frac{A^2} R$ ->eq7

But peak power and average power are same, I think they should not be same, what is the mistake?

Remember $A$ and $\omega$ are given values: $A$ = 170 (presumably volts) and $\omega$ = 377 (presumably rad/s).

For a simple resistor (V and I in phase), there are 3 forms of the power formula:
$P = VI$, $P = I^2 R$ and $P = \frac {V^2} {R}$

Using the 3rd formula let's you answer part a) straight away.
$P(t) = \frac{A^2\sin^2(\omega t)} R$ which you correctly got.
I don’t think you need to simplify further (using the double angle). But I expect you are meant to use the given values for $A$ and $\omega$, because the question does not use the symbols $A$ and $\omega$. If I were answering I would simply write:
a) $P(t) = \frac {V(t)^2}{R} = \frac{170^2\sin^2(377t)} {10} = 2890sin^2(377t)~watts$

The power changes continuously during each cycle. The peak power is simply the maximum value of instantaneous power. Using the part a) answer allows you to answer part b) without any calculation.

For part c), your 1st equation is wrong. You have
$p_{avg} = \frac{\int_0^{2\pi} A^2(1+\cos(2\omega t))dt} {2R\pi}$
But the denominator is $2RT = 2R(2\pi) = 4R\pi$

Also note, a less messy approach is to move all constant factors outside the integral. Tis gives less cluttered working.

EDIT: Note the period T is in fact $\frac {2\pi} {\omega}$ so integrating from 0 to $2\pi$ s not correct - but ends up giving the correct answer 'by accident'.