Electric Current as a function of time

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SUMMARY

The discussion centers on calculating the total charge passing through a conductor with a time-varying electric current described by the equation I(t) = 100 sin(120πt). The correct approach involves using the integral Q = ∫ I dt from t = 0 to t = 1/240 s. The integral calculation yields a result of -0.265, but the definite integral evaluates to 0 due to the behavior of the cosine function at the specified limits.

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  • Understanding of electric current and charge relationships (I = dQ/dt)
  • Knowledge of definite integrals in calculus
  • Familiarity with trigonometric functions and their properties
  • Ability to use a scientific calculator for trigonometric calculations
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  • Study the properties of sine and cosine functions in relation to integrals
  • Learn how to perform definite integrals involving trigonometric functions
  • Explore the concept of electric charge and current in physics
  • Practice solving problems involving time-varying currents
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jmuduke
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Homework Statement


8. An electric current in a conductor varies with time according to the expression
I(t) = 100 sin (120*pi*t), where I is in amperes and t is in seconds. What is the total charge passing a given point in the conductor from t = 0 to t = 1/240 s?




Homework Equations






The Attempt at a Solution


I have attempted to substitute the values of t into the equation and use the difference, but I do not feel that was the correct way. Next, I attempted to perform a definite integral, but I get 0 for both numbers.
 
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welcome to pf!

hi jmuduke! welcome to pf! :smile:
jmuduke said:
I(t) = 100 sin (120*pi*t), where I is in amperes and t is in seconds. What is the total charge passing a given point in the conductor from t = 0 to t = 1/240 s?

I attempted to perform a definite integral, but I get 0 for both numbers.

yes, I = dQ/dt, so Q = ∫ I dt

it's only over 90°, so you shouldn't get 0 for both limits :confused:

show us what you did :smile:
 
Thanks for the reply Tim!

I calculated the integral and got -(5 cos(120*pi*t))/6*pi

Originally, my calculator was set in radians, so that could have been why I got 0 for both limits. I changed it to degrees and got -0.265 for both limits then, but that result in the definite integral being 0, correct?
 
hi jmuduke! :smile:

(just got up :zzz:)
jmuduke said:
I calculated the integral and got -(5 cos(120*pi*t))/6*pi

… got -0.265 for both limits then, but that result in the definite integral being 0, correct?

yes, cos(0) = 1, so that's correct for the t = 0 limit :smile:

but for t = 1/240, cos(120πt) = cos(π/2) = cos90° = 0 :wink:
 

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