Kirchhoff's Rules: Solving Homework Questions on Circuit Currents

  • Thread starter Thread starter Roze
  • Start date Start date
  • Tags Tags
    Rules
AI Thread Summary
The discussion focuses on applying Kirchhoff's Rules to analyze a circuit and find the currents. The confusion arises from the choice of resistors in the voltage equations, specifically why the 9.0-ohm resistor is included instead of the 5.0-ohm resistor. It is clarified that the direction of current flow determines whether voltage is considered positive or negative, with voltage dropping across resistors and gaining across batteries. The need for a third equation, based on the junction rule, is emphasized to account for the relationship between the currents. The question of why Kirchhoff's Rules are used for this circuit, despite the potential for simpler methods, is also raised.
Roze
Messages
14
Reaction score
0

Homework Statement


Find the currents in the circuit shown in the figure attached.

Homework Equations


I actually have the solution worked out but I don't understand why something was done. So they chose to travel the bottom loop and the top loop clockwise in both cases. So they set up the equation:

sum deltaV = Vbattery + V4.0 + V9.0

What I don't understand is why they chose to add the V9.0, rather than the V5.0, or both. How do I know that this is the resistor you use for this equation? I understand you're traveling the loop clockwise, but I don't understand why the 9.0 resistor is used rather than the 5.0 or their equivalent resistance.

Also, when they plug in the numbers for this equation they give:

sum deltaV = 6.0V -4.0I2 - 9.0I3

Why are the 4.0 and 9.0 components negative? Is it because they're on the other side of the battery?

Similarly, for the top loop they give the equation:

sum deltaV = -5.0I2 + 9.0I3

Why in this case is the 5.0 negative but the 9.0 positive?

The Attempt at a Solution



See above.

Apologies for the terrible picture.
 

Attachments

  • 18.4.gif
    18.4.gif
    5.4 KB · Views: 478
Physics news on Phys.org
You have to have the concept of a "voltage drop" across a resistor. As the current flows, the flowing charges have more energy on the left side of the resistors (in this image) than they do on the right sides. It helps to draw arrows above each resistor to show the direction that current (conventional current, I) is going. THis way you can tell if you you are dropping voltage ("negative," is going with the arrow) or gaining voltage (going against the arrow).

Of course you gain voltage when you go "with" the current through a battery, but you drop voltage when you go with the current through a resistor.

So in the second loop, as you go clockwise, you go with the current through one resistor, but against the current through the other resistor.

And the first loop simply chose the 9 ohm resistor randomly. You have to pick a single path in each loop. The problem would work just as well if they had chosen the 5 ohm resistor.
 
Roze said:
sum deltaV = 6.0V -4.0I2 - 9.0I3

sum deltaV = -5.0I2 + 9.0I3

One of these is incorrect. I am assuming that the current through the 4 ohm resistor is "I1." The current can't be the same through both 4 and 5.

You need a third equation (junction rule) which simply says I1 = I2 + I3.

What I don't understand is, why use Kirchoff's rules for this simple circuit? Just for practice? It would be easier to find the equivalent resistance, then the total current, then find the proportional split of the current through the parallel.
 
Last edited:
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...

Similar threads

Back
Top