How do I simplify this resistor circuit for my homework?

  • Thread starter Thread starter exitwound
  • Start date Start date
  • Tags Tags
    Circuit Resistors
AI Thread Summary
To simplify the resistor circuit, the bottom branch combines to 60 Ohms. The top right resistor is shorted out and can be disregarded. The remaining resistors consist of 30 Ohms in parallel with another 30 Ohms and the combined 60 Ohms from the bottom. The equivalent resistance is calculated using the formula for parallel resistors, resulting in 12 Ohms. This approach clarifies the circuit's simplification process for the homework assignment.
exitwound
Messages
291
Reaction score
1

Homework Statement



Capture2.JPG


Homework Equations



Resistors in Series = R1 + R2 + R3... = Req
Resistors in Parallel = 1/R1 + 1/R2 + 1/R3 ... = 1/Req

The Attempt at a Solution



Studying for a test. This sample exam went pretty well. I am having trouble with 3 out of the 20. I can't figure this one out.

The bottom branch can be combined to be Req = 30 + 30 = 60 Ohms

From there, I can't seem to find a way to combine the rest. I have the answer (B) but I can't collapse the circuit.
 
Physics news on Phys.org
The top right resistor is shorted out, so it has no effect. Imagine that it's erased from the diagram.
The center point and point b are the same point.
You have 30 in parallel with 30 in parallel with (30+30).
1/[(1/30) + (1/30) + (1/{30+30})] = 12 ohms.
 
Last edited:
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top