Problem finding the direction of current (electricity)

AI Thread Summary
The discussion revolves around understanding the direction of current in an electrical circuit, specifically I1, I2, and I3. It highlights that I1 equals the sum of I2 and I3, and that using Kirchhoff's Voltage Law (KVL) requires assuming current directions initially. If a calculated current is negative, it indicates the assumed direction was incorrect, which can be easily corrected. The conversation also contrasts KVL with Kirchhoff's Current Law (KCL), noting that KCL does not present the same directional assumption issue. Overall, the participants emphasize the self-correcting nature of circuit analysis methods.
AXiMe
Messages
2
Reaction score
0
1. Find I1, I2 and I3, see attached file.



2. I1 = I2 + I3, U = R * I



3. Can't seem to understand why the current is going the way it is at I3 rather than the other way.
 

Attachments

  • fysikkoppgave.png
    fysikkoppgave.png
    4.9 KB · Views: 476
Physics news on Phys.org
AXiMe said:
1. Find I1, I2 and I3, see attached file.



2. I1 = I2 + I3, U = R * I



3. Can't seem to understand why the current is going the way it is at I3 rather than the other way.

When you solve for the currents, it may well turn out that I3 turns out to have a negative sign, which would mean that it actually flows the other way. When starting out KVL loop equations, you just assume some direction for each branch current, and then solve. If some of them turn out negative, you can go back and reverse the arrow and change the answer to positive. No big deal.

Welcome to the PF, BTW!
 
Thanks! Albeit, I still don't understand why it would come out negative, even though the answer says so...
 
AXiMe said:
Thanks! Albeit, I still don't understand why it would come out negative, even though the answer says so...

I didn't work through the problem you posted, but this is just a general part of using KVL to solve circuits. You have to assume some direction for the current in each leg before you can solve the equations. Once you solve the equations, if the current comes out with a negative sign, then that just means that the initial guess at current direction was wrong. No big deal, the process is self-correcting.

When you use KCL equations, you don't have this issue, because you just label nodes with some voltages like V_1, V_2, etc. When you solve the simultaneous equations, you get whether they are postive or negative (or changing) voltages as part of the solution.
 
AXiMe said:
Thanks! Albeit, I still don't understand why it would come out negative, even though the answer says so...

Oh, and don't call me Albeit. I prefer Shirley. :biggrin:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top