Are my KVLs incorrect for this circuit?

  • Thread starter Thread starter pyroknife
  • Start date Start date
AI Thread Summary
The discussion centers on the correctness of Kirchhoff's Voltage Law (KVL) equations for two loops in a circuit. The first loop equation presented is 12 - 12*i1 - 10(i1 - i2) - 6*i1 = 0, while the second loop equation is contested, with the user proposing -30(i2 - i1) + 10(i2 - i1) = 0, while others suggest it should be -30(i2 - i1) - 10(i2 - i1) = 0. Clarifications are made regarding the sign conventions for voltage drops and rises based on the direction of current flow. The consensus leans towards the alternative solution for loop 2, emphasizing the importance of consistent sign usage when applying KVL. Accurate application of KVL is crucial for solving circuit problems effectively.
pyroknife
Messages
611
Reaction score
4
I attached the problem. I'm writing the KVL for each separate loop. For my purposes, I will only write the first two.

i1=1st loop (one on the left) current
i2=2nd loop current (loop in the middle)

Loop 1: 12-12*i1-10(i1-i2)-6*i1=0
Loop 2: -30(i2-i1)+10(i2-i1)=0

The solutions differ with mine however that for loop 2 they have -30(i2-i1)-10(i2-i1)=0

Am I wrong?
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    4 KB · Views: 350
Physics news on Phys.org
pyroknife said:
Loop 2: -30(i2-i1)+10(i2-i1)=0

The solutions differ with mine however that for loop 2 they have -30(i2-i1)-10(i2-i1)=0

Am I wrong?
I agree more with the solution version -- apart from the fact that it's (i2-i3) that goes with the 30 ohm resistor.

The way I remember it is: when traversing around the loop, when going in the same direction as the current then it's a voltage drop, therefore you get a "-" sign. When going against the current direction it's a voltage rise, and therefore you get a "+" sign.

So, if we agree that we traverse the middle loop clockwise, and the currents are all defined as clockwise as well, then it should be "-" for all the i2 terms and "+" for i1 and i3 in the middle-loop equation.

Hope that helps.
 
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