Finding an Unknown Resistance of circuit

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Homework Help Overview

The discussion revolves around finding an unknown resistance in a circuit where the ammeter reading remains constant regardless of the switch positions. The known resistances are R1 = 103 Ω, R2 = 279 Ω, and R3 = 46.0 Ω.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to analyze the circuit with switches both open and closed, equating the resistances in both scenarios based on the constant current through R1. They explore the implications of the ammeter reading and the voltage across R1.

Discussion Status

Some participants have provided insights into determining the current through R1 when the switches are open and how to visualize the circuit when the switches are closed. There is an ongoing exploration of how current splits in the circuit and the application of Kirchhoff's laws.

Contextual Notes

Participants are considering the effects of switch positions on the circuit's behavior and are questioning how to accurately account for current distribution and voltage drops in their calculations.

SMA777
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Homework Statement



In the circuit in the figure below, the reading of the ammeter is the same when both switches are open and when both switches are closed. What is the unknown resistance R? (Let R1 = 103 Ω, R2 = 279 Ω, and R3 = 46.0 Ω.)

25-p-084-alt.gif


Homework Equations



Adding Resistance in Series: R1 + R2 = Req
Adding Resistance in Parallel: 1/R1 + 1/R2 = 1/Req
V = IR ... I = V/R

The Attempt at a Solution



I believe you have to find the resistance in both cases, switch open and closed, and then set them equal because I = V/R is equal in both cases, so R must be equal, since V doesn't change (always 1.5).

For switch OPEN, I got: Req,open = R2 + R1 + R3
For switch CLOSED, I got: Req, closed = R2 + (1/R1 + 1/Runknown)^-1

I set these equal and solved for Ruknown and got 316.8, but that isn't right. Any tips? Thank you!
 
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SMA777 said:

Homework Statement



In the circuit in the figure below, the reading of the ammeter is the same when both switches are open and when both switches are closed. What is the unknown resistance R? (Let R1 = 103 Ω, R2 = 279 Ω, and R3 = 46.0 Ω.)

25-p-084-alt.gif


Homework Equations



Adding Resistance in Series: R1 + R2 = Req
Adding Resistance in Parallel: 1/R1 + 1/R2 = 1/Req
V = IR ... I = V/R

The Attempt at a Solution



I believe you have to find the resistance in both cases, switch open and closed, and then set them equal because I = V/R is equal in both cases, so R must be equal, since V doesn't change (always 1.5).

For switch OPEN, I got: Req,open = R2 + R1 + R3
For switch CLOSED, I got: Req, closed = R2 + (1/R1 + 1/Runknown)^-1

I set these equal and solved for Ruknown and got 316.8, but that isn't right. Any tips? Thank you!

What remains the same in both cases is the current through R1 (as measured by the ammeter). So I'd suggest first determining what that current is when the switches are both open. Also determine the voltage that will appear across R1.

Next draw a sketch of the circuit that results when the switches are closed (are any components bypassed and hence removed from the circuit?). Add the voltage and current you determined above to the sketch -- the current through the ammeter and hence the voltage across R1 should remain the same as before. Remember that an ideal ammeter has no resistance, hence no voltage drop will appear across it. Can you find the remaining currents?
 
Hi SMA777! :smile:
SMA777 said:
For switch CLOSED, I got: Req, closed = R2 + (1/R1 + 1/Runknown)^-1

I set these equal …

No, the Req you found (for switches closed) gives you I = V/Req for the current next to the battery …

that current will split before going through the ammeter, won't it? :wink:
 
Oh, I see what you mean about how it splits up! Ok, got it. My question is, how do I know... how it splits up? IE how much goes to Runknown and how much to R1? Is the sum of their currents = total?
 
(just got up :zzz: …)
SMA777 said:
… Is the sum of their currents = total?

Yes, from KCL.

Then use KVL for the loop containing just R1 and R. :smile:
 

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