Finding an Unknown Resistance of circuit

AI Thread Summary
The discussion revolves around determining the unknown resistance (R) in a circuit where the ammeter reading remains constant regardless of whether the switches are open or closed. The user initially calculates the equivalent resistance for both scenarios but arrives at an incorrect value for R. Key insights include the need to analyze the current through R1 when the switches are open and how it splits when the switches are closed. The correct approach involves applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to find the current distribution and voltage across the components. Ultimately, understanding how the current splits between R1 and Runknown is crucial for solving the problem accurately.
SMA777
Messages
11
Reaction score
0

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!
 
Physics news on Phys.org
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:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top