Wrong value for the voltage across a capacitor

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The discussion revolves around calculating the voltage across a 100 microfarad capacitor given a specific current function. The user initially computed the voltage as -2.87 * 10^(-5) V, while the textbook states it should be 93.14 mV. A key point of confusion arose from the use of radians versus degrees in the cosine function, which led to incorrect calculations. It was clarified that the negative sign in the voltage expression should be considered, as the maximum value of cosine is 1, ensuring the final answer is positive. The consensus confirms that the book's answer is correct, highlighting the importance of proper unit conversion and trigonometric function usage.
AfterSunShine
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Homework Statement


The current through a 100 microfarad is i(t) = 50 sin(120*pi*t) mA
Calculate the voltage across it at t = 1 ms. Consider zero initial voltage.

Homework Equations


v(t) = (1/c) integral of i(t) from t = 0 to t = t

The Attempt at a Solution


Am getting v(t) = - 1.326 ( cos(120*pi*t) - 1 ) V
which gives v(1*10^-3) = -2.87 * 10^(-5) V
But book is saying the voltage should be 93.14 mV

Can any confirm if book answer is correct?
I did the problem like 100 times, and still same value. I considered that given current in mA so I multiplied by 10^-3, and considered the 10^-6 from micro farad. integral of sin(120*p*t) is -cos(120*pi*t) / (120*pi)
I have no idea where is my mistake.
 
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The book's answer is correct, but so is most of your calculations. When you computed ## \cos(.377) ## , did you use radians or degrees? You need to use radians. ## \\ ##Edit: And yes, I tried the arithmetic with degrees, and I got very close to your (incorrect) answer, but I get +2.87 E-5, with a "+" sign.
 
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So i substitute pi with 180.
And yes you are correct, while calculating final answer i forget to include the "-" sign
I have negative sign in my v(t), and there will (cos(something) - 1) which will give minus for sure as maximum value for cosine is 1, with the negative in v(t), answer must be positive
Thanks!
 
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