Calculating Resistance in a Parallel Circuit without Voltage Information

Click For Summary

Homework Help Overview

The discussion revolves around calculating the resistance of a resistor in a parallel circuit, given the total current and the current through one of the resistors. The original poster expresses confusion due to the absence of voltage information, which is typically necessary for such calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between total current and individual currents in a parallel circuit. There are attempts to apply Ohm's law and the formula for equivalent resistance, but uncertainty remains regarding which values to use due to missing information about the other resistors.

Discussion Status

Participants are actively engaging with the problem, suggesting various methods to find the total voltage and resistance. Some guidance has been offered regarding the use of equivalent resistance and Ohm's law, though multiple interpretations of the approach are still being discussed.

Contextual Notes

There is a noted lack of information about the resistances of R1 and R2, which complicates the calculations. The discussion also highlights the challenge of working without explicit voltage data.

Aleena753
Messages
7
Reaction score
0

Homework Statement


Three resistors are connected as shown in the diagram. If the current from the power supply is 250 mA and the current through R3 is 54 mA, what is the resistance of R3?
FullSizeRender (2).jpg


Homework Equations


1/Req= 1/R1 + 1/R2 +1/R3

V=IR

The Attempt at a Solution



First I used the parallel circuit formula of 1/Req= 1/R1+ 1/R2 + 1/R3
but there are two variables (Req and R3) so I'm not sure how to use that. Also, there's no voltage provided so I'm a bit confused as how to go from here. Any help would be appreciated :)
 
Physics news on Phys.org
You can see that the resistors are in parallel, meaning voltage across each resistor is equal to the battery voltage.
Aleena753 said:
current from the power supply is 250 mA
Aleena753 said:
and the current through R3 is 54 mA,
Could you calculate voltage of the battery from this information?
Hint: Ohm's law and resistances in parallel.
 
Last edited:
  • Like
Likes   Reactions: Aleena753
I had an idea for finding the solution. I was thinking that I could find the total V and then divide it by the total current to find Req and then use that Req in the parallel circuit formula (1/Req= 1/R1 + 1/R2 + 1/R3) to find R3. As for the total voltage of the battery, I know to use V=IR but I'm not sure which numbers to use, since the current of the two known resistors is unknown.
 
Aleena753 said:
since the current of the two known resistors is unknown.
Well, you can combine the two known resistances by finding their equivalent resistance. You can then calculate current through the equivalent resistance and apply Ohm's law.
 
So the current for the equivalent resistance of R1 and R2 would be the total current of the circuit with the current of the third resistor subtracted right?
 
Aleena753 said:
So the current for the equivalent resistance of R1 and R2 would be the total current of the circuit with the current of the third resistor subtracted right?
Right.
 
I had an idea for finding the solution. I was thinking that I could find the total V and then divide it by the total current to find Req and then use that Req in the parallel circuit formula (1/Req= 1/R1 + 1/R2 + 1/R3) to find R3. As for the total voltage of the battery, I know to use V=IR but I'm not sure which numbers to use, since the current of the two known resistors is unknown.[/QUOTE]
cnh1995 said:
You can see that the resistors are in parallel, meaning voltage across each resistor is equal to the battery voltage.Could you calculate voltage of the battery from this information?
Hint: Ohm's law and resistances in parallel.
cnh1995 said:
Right.
So I found the voltage and then divided it by the third current (54mA) to find the resistance of the third resistor. I currently have what I believe is the right answer of 333 ohms.
 
Aleena753 said:
I currently have what I believe is the right answer of 333 ohms.
I'm getting 334.something ohms. But your method is correct.
 
cnh1995 said:
I'm getting 334.something ohms. But your method is correct.
Alright thank you for your help : )
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K