Wrong value for the voltage across a capacitor

In summary, the conversation involves calculating the voltage across a 100 microfarad capacitor at t = 1 ms, given the current i(t) = 50 sin(120*pi*t) mA and zero initial voltage. The textbook answer is 93.14 mV, while the OP's calculations result in -2.87 * 10^(-5) V. The discrepancy is due to the use of degrees instead of radians in the computation of cos(120*pi*t). After correcting this mistake, the answer matches with the textbook.
  • #1

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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  • #2
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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  • #3
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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