Circuit: Total Energy into a component over interval

Click For Summary
SUMMARY

The discussion focuses on calculating the total energy absorbed by an electric component over a specified time interval using the voltage function v(t) = 20cos(120πt) Volts and the current function i(t) = -4sin(120πt) amps. The power is defined as the product of current and voltage, leading to the expression for energy ΔE from time t1 to t2 as ΔE_{t1→t2} = (1/6π)[cos(240πt)]_{t1}^{t2}. A specific case is evaluated for t2 = t1 + 1/15, prompting a discussion on utilizing trigonometric identities to simplify the evaluation of the cosine function.

PREREQUISITES
  • Understanding of electrical power calculations
  • Familiarity with trigonometric identities and periodic functions
  • Knowledge of integration techniques in calculus
  • Basic concepts of complex exponentials in relation to trigonometric functions
NEXT STEPS
  • Study the derivation of energy from power in electrical circuits
  • Learn about the application of trigonometric identities in energy calculations
  • Explore complex exponential representations of trigonometric functions
  • Investigate the periodic properties of cosine functions in energy analysis
USEFUL FOR

Students in electrical engineering, physics enthusiasts, and anyone involved in circuit analysis and energy calculations in electrical components.

Saladsamurai
Messages
3,009
Reaction score
7

Homework Statement



The voltage at terminal a relative to terminal b of an electric component is given by v(t) = 20\cos(120\pi t) Volts and the current into a is i(t) = -4\sin(120\pi t) amps. Find 1) the total energy that flows into component from time t1 to t2 and 2) in particular find the energy absorbed when t2 = t1 + 1/15.


Homework Equations



Power = i(t)*v(t)

The Attempt at a Solution



So I set Power = dE/dt = i(t)*v(t) and integrated to yield the final expression

\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos(240\pi t)\right]_{t_1}^{t_2}\qquad(1)

I believe that this expression takes care of part 1). However, for part 2), I am unclear on how to evaluate the expression from t1 to t1 + 1/15. This becomes:

\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos(240\pi t_1+1/15) - \cos(240\pi t_1) \right]

Is there some sort of trig trick I a can use to evaluate this? Or somehow use the fact the a cosine function is periodic?

Just need a hint here :smile:
 
Physics news on Phys.org
Saladsamurai said:
\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos({\color{red}240\pi t_1+1/15}) - \cos(240\pi t_1) \right]
Maybe it's just a typo within LaTeX, but you didn't multiply through the 240π quite correctly.
Is there some sort of trig trick I a can use to evaluate this? Or somehow use the fact the a cosine function is periodic?

Just need a hint here :smile:
Here's a hint: Note that

\cos (\omega t + \theta) = \frac{e^{j(\omega t + \theta)} + e^{-j( \omega t + \theta)}}{2}

Try to express your equation exponentially. You'll be able to pull out a complex exponential (that's not a function of t1), and functions as a complex constant. Recalling that e^{j \theta} = \cos \theta + j \sin \theta [/tex], the whole thing can be reduced in this particular problem.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
23
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K