Solving for RMS current of a resistor, help?

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clamatoman
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Homework Statement


A resistor connected across an AC power supply has a current given by I=(1.35A)cos(300t) when connected to a power supply with emf 120 V rms.
Find the RMS current.

Homework Equations


IRMS=Imax/√2

The Attempt at a Solution


IRMS=Imax/√2
IRMS=0.477 A
INCORRECT
Not exactly sure what to do? I am not given any other data besides what i have written above. I do not have the Power, or the Resistance, and in fact will have to solve for those in part B. and C. of this problem. I know I am missing something here, and i have re-read the chapter in my textbook but it is not helping.
 
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gneill said:
How did you establish the value of ##I_{max}## that you used?
I used the value of "I=(1.35A)cos(300t)" they gave as Imax.
 
clamatoman said:
I used the value of "I=(1.35A)cos(300t)" they gave as Imax.
Okay, but that's not a value. What value (single number) did you use?
 
gneill said:
Okay, but that's not a value. What value (single number) did you use?
I used (1.35A)cos(300t) = 0.675
I think what I am missing is what the "t" component means?
 
clamatoman said:
I used (1.35A)cos(300t) = 0.675
That is incorrect for a couple of reasons. First, the expression cos(300t) is a function of time t. Second, the "300" in the expression would have units of radians per second, not degrees per second.

Read the maximum value from the coefficient: 1.35 A. The cosine function makes the current vary between the bounds -1.35 A and +1.35 A over time.
 
gneill said:
That is incorrect for a couple of reasons. First, the expression cos(300t) is a function of time t. Second, the "300" in the expression would have units of radians per second, not degrees per second.

Read the maximum value from the coefficient: 1.35 A. The cosine function makes the current vary between the bounds -1.35 A and +1.35 A over time.
Ahh Okay.
So.
Imax=1.35 A
IRMS=1.35/√2=.955 A
Excellent, thank you.