Calculating Power Dissipated in 20 Ohm Resistor: I = 12 sin(250t)

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SUMMARY

The power dissipated in a 20 Ohm resistor with an electric current described by I = 12 sin(250t) is calculated using the formula P = I^2 * R. The correct expression for power is P = 2880 sin^2(250t). To find the average power dissipated as heat, one must integrate this expression over a time period and divide by that period to obtain the root mean square (RMS) value. This approach ensures accurate calculation despite the lack of efficiency information or specific time values provided in the problem.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with power equations (P = IV, P = I^2 * R)
  • Knowledge of RMS calculations in electrical engineering
  • Basic calculus for integration over time
NEXT STEPS
  • Learn about RMS value calculations for alternating currents
  • Study the integration of trigonometric functions over a period
  • Explore the concept of power factor in AC circuits
  • Review the principles of energy dissipation in resistive components
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Electrical engineering students, physics learners, and professionals involved in circuit design and analysis who need to calculate power dissipation in resistive loads.

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Homework Statement


An electric current described by I = 12 sin(250t) flows through a 20 ohm resistor. Calculate the power dissipated as heat.


Homework Equations


V=IR
P=IV
P = I^2*R


The Attempt at a Solution


Using the last equation, P = I^2*R
P = 2880 sin^2(250t)
Is that the best answer I can give? The question gives no information about efficency so I'm not sure if I'm forgetting an equation or something. It doesn't give a value for time either.
 
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I think you misinterpreted the problem. The problem asks for the power dissipated as heat, not the power in the element.
 
You need to integrate over a time period then divide by the period to get the root mean square (rms) value.
 

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