1006 Words5 Pages

Trigonometry
Problem #1
The electricity supplied to your house is called alternating current (AC) because the current varies sinusoidally with time. The voltage which causes the current to flow also varies sinusoidally with time. Bothe the current and the voltage have a frequency of 60 cycles per second which corresponds to a period of 1/60 of a second, but current and voltage have different phase shifts. We will be determining equations for the voltages and the current as functions of time based upon the equationf(t)=Acos[B(x-C)]+D.
Under the assumption that current and voltage are both sinusoidal functions of time (t, in seconds) and both have the same frequency of 60 cycles per second (period=1/60), let C be the current in Amperes (Amps) with a maximum current of 5 Amps at zero seconds (t=zero). Let V be the voltage that is “leading” the current (‘leading’ corresponding to a negative phase shift) by .003 seconds, meaning voltage reaches its maximum of 180 volts .003 sec before the current reaches its maximum. Both functions begin at zero- there are no phase shifts.
Question 1, Part A. Current as a function of time: Based upon the given information, we will be determining the Amplitude (A), period (and thus, B), and the phase shift (C). Given information states there is not a phase shift, therefore the value of D is zero and does not need to be recorded throughout the problem. The Amplitude (A) or maximum for the current has been given: 5 Amps. This indicates that the range will span from -5 to 5 on the y-axis. It is also given that at t=zero (time=zero) the current is equal to 5 Amps, meaning there is a negative phase shift (or phase shift to the left). A=5 B is calculated by the equation B=2π/( 1/60) , where p (period length)= 1/60 or the length of one full cycle found from the frequency which is 60cycles per second. So p is

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