Circuit Segment (Two sources with a resistor in between)

AI Thread Summary
The discussion focuses on calculating the voltage difference Va−Vb in a circuit with two voltage sources and a resistor. The current is given as 4 A, with resistors R1 and R2 both at 6 ohms, and voltage sources V1 at 16 V and V2 at 6 V. The solution involves using Ohm's Law (V=IR) and Kirchhoff's Voltage Law (KVL) to account for potential drops across the resistors and the effects of the voltage sources. The final answer is determined to be 58 V, with a hint suggesting a relationship between the contributions of the voltage sources. The discussion emphasizes the importance of correctly assigning potential drops and the direction of current flow in the calculations.
Bluestribute
Messages
192
Reaction score
0

Homework Statement


If a current I=4 A exists between points a and b, R1=6 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, R2=6 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, V1=16 V, and V2=6 V,the voltage difference Va−Vb is

Homework Equations


V=IR

The Attempt at a Solution


Well, I do V=IR for the first section up to source V1 (so V=4(6)), but once I get in between, I don't know how to deal with the inside having two voltage sources bombarding it. The answer is 58 V (it's multiple choice), but how?
 

Attachments

  • circuitSegment.gif
    circuitSegment.gif
    1.4 KB · Views: 641
Last edited by a moderator:
Physics news on Phys.org
From the answer, you can guess the physics. Hint: 24 + 24 + x = 58?
What does the value of x imply the direction of the potential difference caused by the voltage sources?
 
You have the current and its direction. So you can assign potential drops across the resistors. Pencil them in on your diagram. Then do a "KVL walk" from b to a summing up the potential changes as you go.
 
gneill said:
You have the current and its direction. So you can assign potential drops across the resistors. Pencil them in on your diagram. Then do a "KVL walk" from b to a summing up the potential changes as you go.
There we go! I thought you had to do something with the two sources going towards the resistor.

I just went from A to B doing Kirchoff Law calcs (so going in the direction of the current was a negative, going from positive to negative terminals was negative, and the reverses were positive) and am getting it now. Thank ya!
 
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