Solving for RMS current of a resistor, help?

Click For Summary

Homework Help Overview

The problem involves calculating the RMS current for a resistor connected to an AC power supply, given a specific current function and voltage. The context is within the subject area of electrical circuits and alternating current analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the determination of the maximum current value from the provided function and question the interpretation of the time variable in the cosine function. There is an exploration of how to correctly apply the RMS calculation based on the maximum current.

Discussion Status

The discussion has seen participants clarify the interpretation of the maximum current and the function's time dependency. Some guidance has been offered regarding the correct extraction of the maximum value from the current expression.

Contextual Notes

Participants note the absence of additional data such as power or resistance, which may be relevant for subsequent parts of the problem. There is also an acknowledgment of potential confusion regarding the time variable in the current expression.

clamatoman
Messages
24
Reaction score
0

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.
 
Physics news on Phys.org
How did you establish the value of ##I_{max}## that you used?
 
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
14K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
8
Views
5K
Replies
2
Views
11K
Replies
4
Views
2K