Understanding Voltage Drop in a Circuit

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Homework Help Overview

The discussion revolves around understanding voltage drop in a circuit, specifically analyzing the resistances R1 and R2 in relation to current values and voltage drops. Participants are trying to reconcile their calculations with the provided answers in a textbook.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to deduce the values of R1 and R2 based on current scenarios, questioning the correctness of the textbook answers. There are discussions about typos in the original post and the implications of those errors on the calculations.

Discussion Status

The discussion is ongoing, with participants providing guidance on correcting typos and clarifying the relationships between the resistances and current. There is no explicit consensus on the correct values yet, as multiple interpretations of the problem are being explored.

Contextual Notes

There are noted typos in the original post regarding current values, which some participants are addressing. The original poster's assumptions about the resistances and their roles in defining maximum and minimum current are also under scrutiny.

Lay1
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Homework Statement
Refer to the figure. Assume the green LED drops 2V across it when it is on. The maximum current in the LED is to be set at 10mA and the minimum current is 2mA. Choose values for R1 and R2 to meet these requirements.
Relevant Equations
V=IR
Here is the figure mentioned above.
20230413_182850.jpg

My thinking is that for maximum current, resistance must be minimum. Thus, R1 is not considered which means the voltage drop of R2 is 10V. So, R2 is 1kohm. For minimum, I=2mA, so R1+R2=5kohm, since R2 is 1kohm, R1 must be 4kohm. This is how I deduce. However, the answer shows that R1=5kohm while R2 is 1kohm. I do not understand why it is so. Could someone help me explain why? Thank you as always.
 
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Lay1 said:
For minimum, I=2A, so R1+R2=5kohm, since R1 is 1kohm, R2 must be 4kohm. This is how I deduce.
Change the I=2A typo to I=2mA and I agree with you. I think the solution's answer for the minimum current case is incorrect.
 
Lay1 said:
Thus, R1 is not considered which means the voltage drop of R2 is 10V. So, R1 is 1kohm
R1 is not considered and R1 is 1kohm?
 
cnh1995 said:
R1 is not considered and R1 is 1kohm?
Oh, good catch. Another typo! :smile: Yeah, we need the OP to add a reply with the R1/R2 stuff cleaned up.
 
berkeman said:
Change the I=2A typo to I=2mA and I agree with you. I think the solution's answer for the minimum current case is incorrect.
Thank you for your guidance. And also thank you for your confirmation.
 
cnh1995 said:
R1 is not considered and R1 is 1kohm?
I am really sorry for my careless mistake. And thank your for your correction.
 
Lay1 said:
Thank you for your guidance. And also thank you for your confirmation.
Wait, did you see where there are other typos as well? Please reply with the correct values for R1 and R2, where R2 is the fixed resistance to define the maximum current, and R1 is the variable resistance to define the minimum current when added to R2.
 
berkeman said:
Oh, good catch. Another typo! :smile: Yeah, we need the OP to add a reply with the R1/R2 stuff cleaned up.
I have already fixed. Sorry for your inconvinience.
 
berkeman said:
Wait, did you see where there are other typos as well? Please reply with the correct values for R1 and R2, where R2 is the fixed resistance to define the maximum current, and R1 is the variable resistance to define the minimum current when added to R2.
My answer: R1=4kohms and R2=1kohm
The textbook answer: R1=5kohms and R2=1kohm
 
  • #10
Lay1 said:
I have already fixed. Sorry for your inconvinience.
Oh, okay, I see that now. I still agree with the two values you calculated.

For future reference, it's usually best to add a reply to correct any typos that have been mentioned in your Original Post (OP). Otherwise the replies afterward that point out the typos can be confusing for others reading the thread later. It's okay to leave it as-is for now.
 
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  • #11
berkeman said:
Oh, okay, I see that now. I still agree with the two values you calculated.

For future reference, it's usually best to add a reply to correct any typos that have been mentioned in your Original Post (OP). Otherwise the replies afterward that point out the typos can be confusing for others reading the thread later. It's okay to leave it as-is for now.
I will be careful in future and I will reply as you have explained. Thank you for your understanding and answer.
 
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