Req of this series/parallel circuit?

  • Thread starter Thread starter letsgo
  • Start date Start date
  • Tags Tags
    Circuit
AI Thread Summary
The discussion focuses on calculating the equivalent resistance (Req) of a circuit involving resistors R1, R2, R3, and R4. The initial calculations incorrectly treat R1 and R3 as series, leading to an incorrect Req of 266.7 ohms. A key point raised is the presence of a wire connecting R3, which effectively reduces its resistance to zero when considered in parallel with R3. This results in R3 having no voltage drop, simplifying the circuit significantly. Participants encourage re-evaluating the calculations with the understanding that R3 in parallel with a zero-ohm resistor leads to a new approach for finding the correct Req.
letsgo
Messages
20
Reaction score
0

Homework Statement


What is the Req of this circuit?
The answer should be 150 ohms, but I can't get that.

Homework Equations


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

The Attempt at a Solution


If I treat R1 and R3 as series, I get R13= 200 ohms.
Then if I treat R13 and R2 and parallel I get R123 = 66.7 ohms.
Then if I treat R123 and R4 as series, I get Req = 266.7 ohms.

Also, since the current is 0 (open switch), then the potential difference between points P and Q would also be 0, right?
 
Last edited:
Physics news on Phys.org
A hint..

There appears to be a wire (grey) connected from one end of R3 to the other end. What does that mean for the voltage across R3 or R3 in general?
 
CWatters said:
A hint..

There appears to be a wire (grey) connected from one end of R3 to the other end. What does that mean for the voltage across R3 or R3 in general?
No clue. It's the same as the source's voltage?

Also, did you manage to get Req?
 
The grey wire has a very low resistance, let's assume it's zero Ohms. You therefore have R3 in parallel with a zero Ohm resistor..

For that bit of the circuit..

I/R = 1/R3 + 1/zero

but 1/zero = ∞

so R3 in parallel with zero Ohms = zero Ohms.

The circuit reduces to..
 

Attachments

  • Tk9OP.png
    Tk9OP.png
    1.2 KB · Views: 516
I'll let you have another go at calculating Req.
 
CWatters said:
The grey wire has a very low resistance, let's assume it's zero Ohms. You therefore have R3 in parallel with a zero Ohm resistor..

For that bit of the circuit..

I/R = 1/R3 + 1/zero

but 1/zero = ∞

so R3 in parallel with zero Ohms = zero Ohms.

The circuit reduces to..
Smart, thanks.
 
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...

Similar threads

Back
Top