Voltage drop between two points

AI Thread Summary
To determine the voltage drop between points A and B, the current in the circuit can be calculated using Kirchhoff's Voltage Law (KVL). The current is given by the formula (E2 - E1) / (R1 + R2 + R3 + R4), indicating a clockwise flow due to E2 being greater than E1. The voltage drop can be calculated using V = IR, but confusion arises regarding which resistors to include. The correct approach involves summing the resistances encountered along the path from A to B, which can be either R1 + R2 or the equivalent resistance of the parallel combinations of resistors. Ultimately, understanding the path taken and the direction of current flow is key to accurately determining the voltage drop.
terryds
Messages
392
Reaction score
13

Homework Statement


xzmxqxsnligtgjjl.png


The question is: What's the voltage drop between point A and B ?? (Suppose that E2 > E1)2. The attempt at a solution

First, I need to solve the current flowing in this circuit, which can be done using KVL.
The current flowing is ##\frac{E_2-E_1}{R_1+R_2+R_3+R_4}## (which is clockwise in direction because E2 is bigger than E1)

But,I get confused in determining voltage drop between point A and B
I know that the formula is V = IR
But, I'm confused about the value of the resistor..

Is it R1 + R2 (since from point A to B the current flows from R1 to R2) ?
Or, should the resistor be the parallel series of R1-R2 and R3-R4 that makes an imaginary bridge between A and B ??
 
Physics news on Phys.org
terryds said:
But,I get confused in determining voltage drop between point A and B
I know that the formula is V = IR
But, I'm confused about the value of the resistor..

Is it R1 + R2 (since from point A to B the current flows from R1 to R2) ?
Or, should the resistor be the parallel series of R1-R2 and R3-R4 that makes an imaginary bridge between A and B ??
Your first thought was correct. You determine the potential between two points by taking a "KVL walk" along a path that leads from one point to the other.

You could also have gone the other way around the loop (counterclockwise) from A to B, summing the potential changes along the way. In that case you would be "walking" against the current, so you'd see potential rises as you cross the resistors rather than drops.
 
  • Like
Likes terryds
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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