Calculating Peak Current in a Series RLC Circuit

AI Thread Summary
In a series RLC circuit with a 100 ohm resistor, a 0.10 H inductor, and a 100 microfarad capacitor connected to a 120 V/60 Hz power line, the peak current calculation initially led to confusion. The formula I=V/Z was applied, with Z calculated using the impedance formula. The error arose from not recognizing that the 120 V provided was the RMS value, not the peak value. To find the peak current, the RMS current must be multiplied by the square root of 2. Correcting for this discrepancy yielded the accurate peak current value.
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Homework Statement


A series RLC circuit consists of a 100 ohm resistor, a 0.10 H inductor, and a 100 micro farad capacitor. It is attached to a 120 V/60 Hz power line. What is the peak current, I?


Homework Equations


I=V/Z
Z=sqrt(R^2+(omegaL - (omegaC)^-1))
omega=2pif


The Attempt at a Solution


I=120/sqrt(100^2+(2pi(60)*.1 - (2pi(60)*100x10^-6)^-1)^2)=1.2
The program is saying it's wrong.
I'm confused what I'm doing wrong, exactly. I'm thinking the omega is wrong, but I don't know how I would get omega otherwise.
Thanks for your help.
 
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I eventually figured it out... apparently 120 V is the rms value, which they clearly do NOT indicate what-so-ever... so my value of I was really I rms, and they just wanted it in I. My answer times sqrt(2) gave me the correct answer.
 

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