Consider a circuit where 2 resistors are connected in parallel

AI Thread Summary
In a circuit with two resistors in parallel and one in series, the total resistance can be calculated using two different approaches. The standard formula for total resistance is R_total = R_3 + (1/(1/R_1 + 1/R_2)), while an alternative derived from Kirchhoff's laws gives R_total = 1/(1/(R_1 + R_3) + 1/(R_2 + R_3)). The discrepancy arises because the second method incorrectly assumes the same current flows through both parallel resistors. Correct application of Kirchhoff's laws shows that the total current is the sum of the currents through each loop, leading to consistent results. Understanding the current distribution in the circuit is crucial for accurate calculations.
Oerg
Messages
350
Reaction score
0
This is a conceptual problem

Consider a circuit where 2 resistors are connected in parallel, which are in turn connected to another resistor in series. Let the resistance of the two resistors in parallel be R_1 and R_2. The resistance of the resistor in series is R_3

Normally we would add the resistance like this to find the total resistance

R_{total}=R_3+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}

however, this works out different when i derive the resistance of the circuit anew from kirchhoffs law which would give

R_{total}=\frac{1}{\frac{1}{R_1+R_3}+\frac{1}{R_2+R_3}}

am i missing something? or is it wrong to apply the resistance formula in the case where there is another resistance connected in series?
 
Physics news on Phys.org


Show your work. How did you get your formula? Actually, the formula for parallel and series resistors are derived from Kirchhoff's laws, you should get the same result.

ehild
 


well if you simplify the expressions you would not get the same result, it is immediately apparent that there will be a {{R_3}^2} term in the second expression but not in the first.

The second expression was obtained by considering 2 different loops and combining the resulting current, same as you would for the derivation of the formula for the combination of resistors in parallel.
 
Last edited:


Your 2 loops are: +side of voltage source -> R3 -> R1 -> - side of voltage source
and
+side voltage source -> R3 -> R2 -> - side of voltage source ?

both loops contain R3. The voltage across R3 depends on the currents of both loops.
The voltage across R3 is not equal to R3*(current in one of the loops)

If 2 loops share a voltage source, it isn't a problem, because the voltage across it is
always the same.
 


The picture shows your circuit.
According to Kirchhoff''s Current Law, I3=I1+I2.
According to the loop Law,
UBC+UCA=UBA, that is
E-I3R3-I1R1=0
and E-I3R3-I2R2=0.
The total current is I3, the total voltage is E, and the resultant resistance between A and B is
RAB=E/I3, and it is the same as the first formula.

ehild
 
Last edited:


Ahhh, you even have a diagram, I am so touched. Thanks for your effort on the forums.

I understand now, in my loop, I had assumed that the same current run throughs both resistor and that I could get the resultant current by adding up the current from the other loop. This is of course not equivalent.
 
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