Understanding Kirchoff's Law and Solving Circuit Problems

AI Thread Summary
To calculate the current I3 in a circuit with a voltage source of 3.7V, it's essential to apply Kirchhoff's Voltage Law (KVL) correctly. The equation used, 0.4R2 - 3.7 - 200I3 = 0, led to an incorrect result due to misinterpreting the direction of current flow. When analyzing the circuit, if the current flows down through R3, it results in a potential rise when moving clockwise, which must be accounted for in the calculations. Properly considering the direction of currents will help avoid mistakes in future analyses. Understanding these principles is crucial for accurately solving circuit problems.
Pochen Liu
Messages
52
Reaction score
2

Homework Statement


upload_2018-11-5_12-29-56.png

How do I calculate the current I3?

Homework Equations


I know that I1 = I2 + I3
and that I2 = 0.4

The Attempt at a Solution


So we have a voltage source of 3.7V I would've thought that the equation for a clockwise circuit be
0.4R2 - 3.7 - 200I3 = 0 which gives me an answer of -0.0183 when it should be 0.0187

How am I using the law wrong and how can I prevent doing so in the future?
 

Attachments

  • upload_2018-11-5_12-29-56.png
    upload_2018-11-5_12-29-56.png
    9.2 KB · Views: 542
Physics news on Phys.org
Pochen Liu said:
So we have a voltage source of 3.7V I would've thought that the equation for a clockwise circuit be
0.4R2 - 3.7 - 200I3 = 0 which gives me an answer of -0.0183 when it should be 0.0187

How am I using the law wrong and how can I prevent doing so in the future?
You need to pay attention to the direction of the currents as you do your "KVL walk" around the circuit. For example, ##I_3## is flowing down through ##R_3##, so as you "walk over" ##R_3## going clockwise around the circuit you'll see a rise in potential.
 
gneill said:
You need to pay attention to the direction of the currents as you do your "KVL walk" around the circuit. For example, ##I_3## is flowing down through ##R_3##, so as you "walk over" ##R_3## going clockwise around the circuit you'll see a rise in potential.
Is this because if I go 'walk over' instead of using the voltage it will add? since it's coming from the other direction, therefore I need to change it's mathematical sign around right?
 
Pochen Liu said:
Is this because if I go 'walk over' instead of using the voltage it will add? since it's coming from the other direction, therefore I need to change it's mathematical sign around right?
Right, you need to account for the current directions. If you "walk" in the same direction of the current you'll see a potential drop across the resistor. If you "walk" against the direction of the current you'll see a potential rise.
 
  • Like
Likes Pochen Liu
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